A Projectile Is Thrown Directly Upward and Caught Again. At the Top of Its Path

four Move in Two and Three Dimensions

four.3 Projectile Move

Learning Objectives

By the end of this section, you will be able to:

Use one-dimensional motion in perpendicular directions to analyze projectile move.
Calculate the range, fourth dimension of flying, and maximum height of a projectile that is launched and impacts a flat, horizontal surface.
Observe the time of flying and affect velocity of a projectile that lands at a dissimilar meridian from that of launch.
Summate the trajectory of a projectile.

Projectile motion is the motion of an object thrown or projected into the air, bailiwick only to acceleration every bit a effect of gravity. The applications of projectile motion in physics and engineering are numerous. Some examples include meteors as they enter Earth'due south temper, fireworks, and the motion of any ball in sports. Such objects are called projectiles and their path is called a trajectory. The motion of falling objects as discussed in Motility Forth a Straight Line is a simple one-dimensional type of projectile motion in which there is no horizontal movement. In this department, we consider two-dimensional projectile motion, and our treatment neglects the effects of air resistance.

The near important fact to remember here is that motions along perpendicular axes are independent and thus tin be analyzed separately. We discussed this fact in Displacement and Velocity Vectors, where we saw that vertical and horizontal motions are independent. The key to analyzing ii-dimensional projectile movement is to break it into two motions: one along the horizontal axis and the other along the vertical. (This pick of axes is the most sensible because dispatch resulting from gravity is vertical; thus, in that location is no dispatch along the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal axis the x-axis and the vertical centrality the y-axis. It is not required that nosotros utilise this choice of axes; information technology is merely user-friendly in the case of gravitational acceleration. In other cases we may cull a different set of axes. (Effigy) illustrates the notation for displacement, where we define

$\overset{\to }{s}$

to be the total deportation, and

$\overset{\to }{x}$

and

$\overset{\to }{y}$

are its component vectors along the horizontal and vertical axes, respectively. The magnitudes of these vectors are s, x, and y.

An illustration of a soccer player kicking a ball. The soccer player's foot is at the origin of an x y coordinate system. The trajectory of the soccer ball and its location at 6 instants in time are shown. The trajectory is a parabola. The vector s is the displacement from the origin to the final position of the soccer ball. Vector s and its x and y components form a right triangle, with s as the hypotenuse and an angle theta between the x axis and s. — **Effigy 4.eleven** The full displacement s of a soccer ball at a point along its path. The vector
$\overset{\to }{s}$

has components

$\overset{\to }{x}$

and

$\overset{\to }{y}$

along the horizontal and vertical axes. Its magnitude is s and it makes an bending θ with the horizontal.

To describe projectile motility completely, we must include velocity and acceleration, as well as displacement. We must find their components forth the x- and y-axes. Permit'southward assume all forces except gravity (such every bit air resistance and friction, for example) are negligible. Defining the positive direction to exist upward, the components of acceleration are and so very uncomplicated:

${a}_{y}=\text{−}g=-9.8\,\text{m}\text{/}{\text{s}}^{2}\enspace(-32\,\text{ft}\text{/}{\text{s}}^{2}).$

Because gravity is vertical,

${a}_{x}=0.$

${a}_{x}=0,$

this means the initial velocity in the 10 management is equal to the last velocity in the x direction, or

${v}_{x}={v}_{0x}.$

With these conditions on dispatch and velocity, we can write the kinematic (Equation) through (Equation) for motion in a compatible gravitational field, including the rest of the kinematic equations for a constant acceleration from Move with Constant Acceleration. The kinematic equations for motion in a compatible gravitational field become kinematic equations with

${a}_{y}=\text{−}g,\enspace{a}_{x}=0:$

Horizontal Move

${v}_{0x}={v}_{x},\,x={x}_{0}+{v}_{x}t$

Vertical Motion

$y={y}_{0}+\frac{1}{2}({v}_{0y}+{v}_{y})t$

${v}_{y}={v}_{0y}-gt$

$y={y}_{0}+{v}_{0y}t-\frac{1}{2}g{t}^{2}$

${v}_{y}^{2}={v}_{0y}^{2}-2g(y-{y}_{0})$

Using this gear up of equations, we tin clarify projectile motion, keeping in mind some important points.

Problem-Solving Strategy: Projectile Motion

Resolve the motion into horizontal and vertical components along the x– and y-axes. The magnitudes of the components of displacement
$\overset{\to }{s}$

along these axes are 10 and y. The magnitudes of the components of velocity

$\overset{\to }{v}$

are

${v}_{x}=v\text{cos}\,\theta \,\text{and}\,{v}_{y}=v\text{sin}\,\theta ,$

where 5 is the magnitude of the velocity and θ is its direction relative to the horizontal, as shown in (Figure).
Treat the motion as ii independent one-dimensional motions: one horizontal and the other vertical. Use the kinematic equations for horizontal and vertical move presented earlier.
Solve for the unknowns in the two separate motions: ane horizontal and 1 vertical. Note that the simply common variable between the motions is time t. The trouble-solving procedures here are the same as those for ane-dimensional kinematics and are illustrated in the following solved examples.
Recombine quantities in the horizontal and vertical directions to notice the total deportation
$\overset{\to }{s}$

and velocity

$\overset{\to }{v}.$

Solve for the magnitude and management of the deportation and velocity using

$s=\sqrt{{x}^{2}+{y}^{2}},\enspace\theta ={\text{tan}}^{-1}(y\text{/}x),\enspace{v}=\sqrt{{v}_{x}^{2}+{v}_{y}^{2}},$

where θ is the management of the displacement

$\overset{\to }{s}.$

Figure a shows the locations and velocities of a projectile on an x y coordinate system at 10 instants in time. When the projectile is at the origin, it has a velocity v sub 0 y which makes an angle theta sub 0 with the horizontal. The velocity is shown as a dark blue arrow, and its x and y components are shown as light blue arrow. The projectile's position follows a downward-opening parabola, moving up to a maximum height, then back to y = 0, and continuing below the x axis The velocity, V, at each time makes an angle theta which changes in time, and has x component V sub x and y component v sub y. The x component of the velocity V sub x is the same at all times. The y component v sub y points up but gets smaller, until the projectile reaches the maximum height, where the velocity is horizontal and has no y component. After the maximum height, the velocity has a y component pointing down and growing larger. As the projectile reaches the same elevation on the way down as it had on the way up, its velocity is below the horizontal by the same angle theta as it was above the horizontal on the way up. In particular, when it comes back to y = 0 on the way down, the angle between the vector v and the horizontal is minus s=theta sub zero and the y component of the velocity is minus v sub 0 y. The last position shown is below the x axis, and the y component of the velocity is larger than it was initially. The graph clearly shows that the horizontal distances travelled in each of the time intervals are equal, while the vertical distances decrease on the way up and increase on the way down. Figure b shows the horizontal component, constant velocity. The horizontal positions and x components of the velocity of the projectile are shown along a horizontal line. The positions are evenly spaced, and the x components of the velocities are all the same, and point to the right. Figure c shows the vertical component, constant acceleration. The vertical positions and y components of the velocity of the projectile are shown along a vertical line. The positions are get closer together on the way up, then further apart on the way down. The y components of the velocities initially point up, decreasing in magnitude until there is no y component to the velocity at the maximum height. After the maximum height, the y components of the velocities point down and increase in magnitude. Figure d shows that putting the horizontal and vertical components of figures b and c together gives the total velocity at a point. The velocity V has an x component of V sub x, has y component of V sub y, and makes an angle of theta with the horizontal. In the example shown, the velocity has a downward y component. — **Effigy 4.12** (a) We analyze ii-dimensional projectile movement by breaking it into ii contained one-dimensional motions along the vertical and horizontal axes. (b) The horizontal movement is unproblematic, because
${a}_{x}=0$

and

${v}_{x}$

is a constant. (c) The velocity in the vertical management begins to decrease as the object rises. At its highest point, the vertical velocity is zero. As the object falls toward Earth once more, the vertical velocity increases over again in magnitude but points in the opposite direction to the initial vertical velocity. (d) The x and y motions are recombined to give the total velocity at whatsoever given point on the trajectory.

Instance

A Fireworks Projectile Explodes High and Away

During a fireworks display, a crush is shot into the air with an initial speed of 70.0 m/due south at an bending of

$75.0\text{°}$

in a higher place the horizontal, as illustrated in (Figure). The fuse is timed to ignite the beat just as information technology reaches its highest signal higher up the basis. (a) Calculate the summit at which the shell explodes. (b) How much time passes betwixt the launch of the beat and the explosion? (c) What is the horizontal displacement of the shell when it explodes? (d) What is the total displacement from the indicate of launch to the highest point?

The trajectory of a fireworks shell from its launch to its highest point is shown as the left half of a downward-opening parabola in a graph of y as a function of x. The maximum height is h = 233 meters and its x displacement at that time is x = 125 meters. The initial velocity vector v sub 0 is up and to the right, tangent to the trajectory curve, and makes an angle of theta sub 0 equal to 75 degrees. — **Figure four.thirteen** The trajectory of a fireworks trounce. The fuse is set to explode the crush at the highest point in its trajectory, which is found to be at a height of 233 m and 125 g abroad horizontally.

Strategy

The motion can be broken into horizontal and vertical motions in which

${a}_{x}=0$

and

${a}_{y}=\text{−}g.$

Nosotros can then define

${x}_{0}$

and

${y}_{0}$

to be zero and solve for the desired quantities.

Solution

(a) By "height" nosotros mean the altitude or vertical position y above the starting point. The highest point in whatever trajectory, called the apex, is reached when

${v}_{y}=0.$

Since we know the initial and final velocities, likewise as the initial position, we use the post-obit equation to find y:

${v}_{y}^{2}={v}_{0y}^{2}-2g(y-{y}_{0}).$

Because

${y}_{0}$

and

${v}_{y}$

are both goose egg, the equation simplifies to

$\text{0}={v}_{0y}^{2}-2gy.$

Solving for y gives

$y=\frac{{v}_{0y}^{2}}{2g}.$

Now we must find

${v}_{0y},$

the component of the initial velocity in the y management. It is given by

${v}_{0y}={v}_{0}\text{sin}{\theta }_{0},$

where

${v}_{0}$

is the initial velocity of 70.0 m/s and

${\theta }_{0}=75\text{°}$

is the initial bending. Thus,

${v}_{0y}={v}_{0}\text{sin}\,\theta =(70.0\,\text{m}\text{/}\text{s})\text{sin}\,75\text{°}=67.6\,\text{m}\text{/}\text{s}$

and y is

$y=\frac{{(67.6\,\text{m}\text{/}\text{s})}^{2}}{2(9.80\,\text{m}\text{/}{\text{s}}^{2})}.$

Thus, we accept

$y=233\,\text{m}\text{.}$

Note that because upwardly is positive, the initial vertical velocity is positive, equally is the maximum height, only the acceleration resulting from gravity is negative. Annotation also that the maximum superlative depends just on the vertical component of the initial velocity, then that any projectile with a 67.6-m/south initial vertical component of velocity reaches a maximum height of 233 m (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which exercise reach such heights before exploding. In practice, air resistance is not completely negligible, so the initial velocity would accept to be somewhat larger than that given to reach the same peak.

(b) As in many physics issues, there is more than one way to solve for the time the projectile reaches its highest point. In this instance, the easiest method is to use

${v}_{y}={v}_{0y}-gt.$

Because

${v}_{y}=0$

at the apex, this equation reduces to merely

$0={v}_{0y}-gt$

$t=\frac{{v}_{0y}}{g}=\frac{67.6\,\text{m}\text{/}\text{s}}{9.80\,\text{m}\text{/}{\text{s}}^{2}}=6.90\text{s}\text{.}$

This fourth dimension is also reasonable for large fireworks. If you are able to see the launch of fireworks, notice that several seconds pass before the shell explodes. Another way of finding the fourth dimension is past using

$y\,\text{=}\,{y}_{0}+\frac{1}{2}({v}_{0y}+{v}_{y})t.$

This is left for you as an exercise to complete.

${a}_{x}=0$

and the horizontal velocity is constant, as discussed earlier. The horizontal displacement is the horizontal velocity multiplied by time as given by

$x={x}_{0}+{v}_{x}t,$

where

${x}_{0}$

is equal to zero. Thus,

$x={v}_{x}t,$

where

${v}_{x}$

is the x-component of the velocity, which is given by

${v}_{x}={v}_{0}\text{cos}\,\theta =(70.0\,\text{m}\text{/}\text{s})\text{cos}75\text{°}=18.1\,\text{m}\text{/}\text{s}.$

Time t for both motions is the same, so x is

$x=(18.1\,\text{m}\text{/}\text{s})6.90\,\text{s}=125\,\text{m}\text{.}$

Horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement institute here could exist useful in keeping the fireworks fragments from falling on spectators. When the crush explodes, air resistance has a major effect, and many fragments state directly beneath.

(d) The horizontal and vertical components of the deportation were just calculated, so all that is needed here is to notice the magnitude and management of the deportation at the highest point:

$\overset{\to }{s}=125\hat{i}+233\hat{j}$

$|\overset{\to }{s}|=\sqrt{{125}^{2}+{233}^{2}}=264\,\text{m}$

$\theta ={\text{tan}}^{-1}(\frac{233}{125})=61.8\text{°}.$

Notation that the angle for the displacement vector is less than the initial angle of launch. To come across why this is, review (Figure), which shows the curvature of the trajectory toward the ground level.

When solving (Figure)(a), the expression we establish for y is valid for whatsoever projectile motion when air resistance is negligible. Call the maximum elevation y = h. Then,

$h=\frac{{v}_{0y}^{2}}{2g}.$

This equation defines the maximum height of a projectile to a higher place its launch position and it depends just on the vertical component of the initial velocity.

Check Your Understanding

A rock is thrown horizontally off a cliff

$100.0\,\text{m}$

loftier with a velocity of fifteen.0 one thousand/s. (a) Define the origin of the coordinate system. (b) Which equation describes the horizontal motion? (c) Which equations describe the vertical motility? (d) What is the rock's velocity at the point of impact?

[reveal-answer q="fs-id1165168031779″]Show Solution[/reveal-respond]

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(a) Choose the elevation of the cliff where the rock is thrown from the origin of the coordinate organization. Although it is arbitrary, we typically cull time t = 0 to correspond to the origin. (b) The equation that describes the horizontal motion is

$x={x}_{0}+{v}_{x}t.$

With

${x}_{0}=0,$

this equation becomes

$x={v}_{x}t.$

${y}_{0}=0\,\text{and}\,{v}_{0y}=0,$

these equations simplify greatly to become

$y=\frac{1}{2}({v}_{0y}+{v}_{y})t=\frac{1}{2}{v}_{y}t,\enspace$

${v}_{y}=\text{−}gt,\enspace$

$y=-\frac{1}{2}g{t}^{2},\enspace$

and

${v}_{y}^{2}=-2gy.$

(d) We use the kinematic equations to find the x and y components of the velocity at the indicate of impact. Using

${v}_{y}^{2}=-2gy$

and noting the point of impact is −100.0 m, nosotros find the y component of the velocity at touch on is

${v}_{y}=44.3\,\text{m}\text{/}\text{s}.$

We are given the ten component,

${v}_{x}=15.0\,\text{m}\text{/}\text{s},$

and then we can calculate the full velocity at impact: v = 46.8 m/due south and

$\theta =71.3\text{°}$

below the horizontal.
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Example

Calculating Projectile Move: Tennis Actor

A tennis histrion wins a match at Arthur Ashe stadium and hits a ball into the stands at thirty m/s and at an angle

$45\text{°}$

above the horizontal ((Figure)). On its mode downward, the ball is caught by a spectator x chiliad higher up the point where the ball was hit. (a) Summate the fourth dimension information technology takes the tennis ball to accomplish the spectator. (b) What are the magnitude and management of the ball's velocity at impact?

An illustration of a tennis ball launched into the stands. The player is to the left of the stands and hits the ball up and to the right at an angle of theta equal to 45 degrees and velocity of v sub 0 equal to 30 meters per second. The ball reaches a spectator who is seated 10 meters above the initial height of the ball. — **Effigy 4.14** The trajectory of a tennis ball hit into the stands.

Strategy

Again, resolving this two-dimensional move into two independent one-dimensional motions allows us to solve for the desired quantities. The time a projectile is in the air is governed past its vertical motion solitary. Thus, we solve for t kickoff. While the brawl is rising and falling vertically, the horizontal motility continues at a abiding velocity. This case asks for the final velocity. Thus, nosotros recombine the vertical and horizontal results to obtain

$\overset{\to }{v}$

at last fourth dimension t, determined in the first part of the example.

Solution

(a) While the ball is in the air, it rises and then falls to a concluding position 10.0 m higher than its starting altitude. Nosotros can discover the time for this past using (Effigy):

$y\,\text{=}\,{y}_{0}\,\text{+}\,{v}_{0y}t-\frac{1}{2}g{t}^{2}.$

If nosotros take the initial position

${y}_{0}$

to be zero, then the terminal position is y = ten m. The initial vertical velocity is the vertical component of the initial velocity:

${v}_{0y}={v}_{0}\text{sin}\,{\theta }_{0}=(30.0\,\text{m}\text{/}\text{s})\text{sin}\,45\text{°}=21.2\,\text{m}\text{/}\text{s}.$

Substituting into (Figure) for y gives us

$10.0\,\text{m}=(21.2\,\text{m/s})t-(4.90\,{\text{m/s}}^{\text{2}}){t}^{2}.$

Rearranging terms gives a quadratic equation in t:

$(4.90\,{\text{m/s}}^{\text{2}}){t}^{2}-(21.2\,\text{m/s})t+10.0\,\text{m}=0.$

Use of the quadratic formula yields t = three.79 south and t = 0.54 s. Since the ball is at a height of ten m at 2 times during its trajectory—once on the fashion up and in one case on the style down—we accept the longer solution for the time it takes the ball to achieve the spectator:

$t=3.79\,\text{s}\text{.}$

The time for projectile movement is determined completely by the vertical motion. Thus, whatsoever projectile that has an initial vertical velocity of 21.2 thou/s and lands x.0 m below its starting altitude spends 3.79 s in the air.

(b) Nosotros can find the final horizontal and vertical velocities

${v}_{x}$

and

${v}_{y}$

with the utilise of the consequence from (a). And so, we can combine them to observe the magnitude of the total velocity vector

$\overset{\to }{v}$

and the angle

$\theta$

information technology makes with the horizontal. Since

${v}_{x}$

is constant, we can solve for it at any horizontal location. Nosotros choose the starting point because we know both the initial velocity and the initial angle. Therefore,

${v}_{x}={v}_{0}\text{cos}{\theta }_{0}=(30\,\text{m}\text{/}\text{s})\text{cos}\,45\text{°}=21.2\,\text{m}\text{/}\text{s}.$

The final vertical velocity is given past (Figure):

${v}_{y}={v}_{0y}-gt.$

Since

${v}_{0y}$

was found in part (a) to be 21.ii m/s, we take

${v}_{y}=21.2\,\text{m}\text{/}\text{s}-9.8\,\text{m}\text{/}{\text{s}}^{2}(3.79\,\text{s})=-15.9\,\text{m}\text{/}\text{s}.$

The magnitude of the final velocity

$\overset{\to }{v}$

$v=\sqrt{{v}_{x}^{2}+{v}_{y}^{2}}=\sqrt{{(21.2\,\text{m}\text{/}\text{s})}^{2}+{(\text{−}\,\text{15}.9\,\text{m}\text{/}\text{s})}^{2}}=26.5\,\text{m}\text{/}\text{s}.$

The management

${\theta }_{v}$

is establish using the inverse tangent:

${\theta }_{v}={\text{tan}}^{-1}(\frac{{v}_{y}}{{v}_{x}})={\text{tan}}^{-1}(\frac{21.2}{-15.9})=-53.1\text{°}.$

Significance

(a) As mentioned earlier, the time for projectile movement is determined completely by the vertical motion. Thus, any projectile that has an initial vertical velocity of 21.2 thousand/due south and lands ten.0 grand beneath its starting altitude spends 3.79 south in the air. (b) The negative angle means the velocity is

$53.1\text{°}$

below the horizontal at the point of impact. This outcome is consistent with the fact that the ball is impacting at a point on the other side of the apex of the trajectory and therefore has a negative y component of the velocity. The magnitude of the velocity is less than the magnitude of the initial velocity we expect since it is impacting x.0 m above the launch elevation.

Time of Flight, Trajectory, and Range

Of involvement are the time of flight, trajectory, and range for a projectile launched on a flat horizontal surface and impacting on the same surface. In this case, kinematic equations give useful expressions for these quantities, which are derived in the following sections.

Time of flying

Nosotros can solve for the time of flight of a projectile that is both launched and impacts on a flat horizontal surface by performing some manipulations of the kinematic equations. Nosotros annotation the position and displacement in y must exist zero at launch and at impact on an even surface. Thus, we set the displacement in y equal to zilch and notice

$y-{y}_{0}={v}_{0y}t-\frac{1}{2}g{t}^{2}=({v}_{0}\text{sin}{\theta }_{0})t-\frac{1}{2}g{t}^{2}=0.$

Factoring, we have

$t({v}_{0}\text{sin}{\theta }_{0}-\frac{gt}{2})=0.$

Solving for t gives us

${T}_{\text{tof}}=\frac{2({v}_{0}\text{sin}{\theta }_{0})}{g}.$

This is the fourth dimension of flying for a projectile both launched and impacting on a apartment horizontal surface. (Effigy) does not utilise when the projectile lands at a different elevation than information technology was launched, as we saw in (Figure) of the tennis player hitting the ball into the stands. The other solution, t = 0, corresponds to the time at launch. The fourth dimension of flight is linearly proportional to the initial velocity in the y direction and inversely proportional to thousand. Thus, on the Moon, where gravity is one-sixth that of Earth, a projectile launched with the same velocity equally on World would be airborne six times as long.

Trajectory

The trajectory of a projectile can exist found by eliminating the time variable t from the kinematic equations for capricious t and solving for y(ten). We take

${x}_{0}={y}_{0}=0$

so the projectile is launched from the origin. The kinematic equation for x gives

$x={v}_{0x}t⇒t=\frac{x}{{v}_{0x}}=\frac{x}{{v}_{0}\text{cos}{\theta }_{0}}.$

Substituting the expression for t into the equation for the position

$y=({v}_{0}\text{sin}{\theta }_{0})t-\frac{1}{2}g{t}^{2}$

gives

$y=({v}_{0}\text{sin}\,{\theta }_{0})(\frac{x}{{v}_{0}\text{cos}\,{\theta }_{0}})-\frac{1}{2}g{(\frac{x}{{v}_{0}\text{cos}\,{\theta }_{0}})}^{2}.$

Rearranging terms, we take

$y=(\text{tan}\,{\theta }_{0})x-[\frac{g}{2{({v}_{0}\text{cos}\,{\theta }_{0})}^{2}}]{x}^{2}.$

This trajectory equation is of the form

$y=ax+b{x}^{2},$

which is an equation of a parabola with coefficients

$a=\text{tan}\,{\theta }_{0},\enspaceb=-\frac{g}{2{({v}_{0}\text{cos}\,{\theta }_{0})}^{2}}.$

Range

From the trajectory equation nosotros tin can also observe the range, or the horizontal altitude traveled by the projectile. Factoring (Figure), nosotros take

$y=x[\text{tan}\,{\theta }_{0}-\frac{g}{2{({v}_{0}\text{cos}\,{\theta }_{0})}^{2}}x].$

The position y is nix for both the launch point and the impact point, since we are once more considering only a flat horizontal surface. Setting y = 0 in this equation gives solutions x = 0, corresponding to the launch point, and

$x=\frac{2{v}_{0}^{2}\text{sin}\,{\theta }_{0}\text{cos}\,{\theta }_{0}}{g},$

corresponding to the impact point. Using the trigonometric identity

$2\text{sin}\,\theta \text{cos}\,\theta =\text{sin}2\theta$

and setting x = R for range, nosotros find

$R=\frac{{v}_{0}^{2}\text{sin}2{\theta }_{0}}{g}.$

Notation particularly that (Figure) is valid just for launch and impact on a horizontal surface. We meet the range is directly proportional to the foursquare of the initial speed

${v}_{0}$

and

$\text{sin}2{\theta }_{0}$

, and it is inversely proportional to the dispatch of gravity. Thus, on the Moon, the range would exist 6 times greater than on Earth for the aforementioned initial velocity. Furthermore, nosotros see from the factor

$\text{sin}2{\theta }_{0}$

that the range is maximum at

$45\text{°}.$

These results are shown in (Effigy). In (a) nosotros see that the greater the initial velocity, the greater the range. In (b), we run into that the range is maximum at

$45\text{°}.$

This is true just for conditions neglecting air resistance. If air resistance is considered, the maximum angle is somewhat smaller. It is interesting that the same range is found for ii initial launch angles that sum to

$90\text{°}.$

The projectile launched with the smaller angle has a lower apex than the higher angle, but they both have the same range.

Figure a shows the trajectories of projectiles launched at the same initial 45 degree angle above the horizontal and different initial velocities. The trajectories are shown from launch to landing back at the initial elevation. In orange is the trajectory for 30 meters per second, giving a range R (distance from launch to landing) of 91.8 m. In purple is the trajectory for 40 meters per second, giving a range R of 163 m. In blue is the trajectory for 50 meters per second, giving a range R of 255 m. The maximum height of the projectile increases with initial speed. Figure b shows the trajectories of projectiles launched at the same initial speed of 50 meters per second and different launch angles. The trajectories are shown from launch to landing back at the initial elevation. In orange is the trajectory for an angle of 15 degrees above the horizontal, giving a range R of 128 m. In purple is the trajectory for an angle of 45 degrees above the horizontal, giving a range R of 255 m. In blue is the trajectory for an angle of 75 degrees above the horizontal, giving a range R of 128 m, the same as for the 15 degree trajectory. The maximum height increases with launch angle. — **Figure iv.15** Trajectories of projectiles on level ground. (a) The greater the initial speed
${v}_{0},$

the greater the range for a given initial angle. (b) The effect of initial angle

${\theta }_{0}$

on the range of a projectile with a given initial speed. Note that the range is the same for initial angles of

$15\text{°}$

and

$75\text{°},$

although the maximum heights of those paths are different.

Example

Comparing Golf game Shots

A golfer finds himself in 2 different situations on different holes. On the second pigsty he is 120 m from the green and wants to hit the ball 90 m and allow it run onto the greenish. He angles the shot low to the ground at

$30\text{°}$

to the horizontal to let the ball curlicue after impact. On the fourth pigsty he is 90 m from the light-green and wants to let the ball drib with a minimum amount of rolling after bear on. Here, he angles the shot at

$70\text{°}$

to the horizontal to minimize rolling after impact. Both shots are striking and impacted on a level surface.

(a) What is the initial speed of the brawl at the second hole?

(b) What is the initial speed of the brawl at the fourth hole?

(d) Graph the trajectories.

Strategy

We see that the range equation has the initial speed and bending, so we can solve for the initial speed for both (a) and (b). When nosotros take the initial speed, we can use this value to write the trajectory equation.

Solution

(a)

$R=\frac{{v}_{0}^{2}\text{sin}\,2{\theta }_{0}}{g}⇒{v}_{0}=\sqrt{\frac{Rg}{\text{sin}\,2{\theta }_{0}}}=\sqrt{\frac{90.0\,\text{m}(9.8\,\text{m}\text{/}{\text{s}}^{2})}{\text{sin}(2(70\text{°}))}}=37.0\,\text{m}\text{/}\text{s}$

(b)

(c)

$\begin{array}{cc} y=x[\text{tan}\,{\theta }_{0}-\frac{g}{2{({v}_{0}\text{cos}\,{\theta }_{0})}^{2}}x]\hfill \\ \text{Second hole:}\,y=x[\text{tan}\,70\text{°}-\frac{9.8\,\text{m}\text{/}{\text{s}}^{2}}{2{[(37.0\,\text{m}\text{/}\text{s)(}\text{cos}\,70\text{°})]}^{2}}x]=2.75x-0.0306{x}^{2}\hfill \\ \text{Fourth hole:}\,y=x[\text{tan}\,30\text{°}-\frac{9.8\,\text{m}\text{/}{\text{s}}^{2}}{2{[(31.9\,\text{m}\text{/}\text{s)(}\text{cos}30\text{°})]}^{2}}x]=0.58x-0.0064{x}^{2}\hfill \end{array}$

(d) Using a graphing utility, we tin compare the two trajectories, which are shown in (Figure).

Two parabolic functions are shown. The range for both trajectories is 90 meters. One shot travels much higher than the other. The higher shot has an initial velocity of 37 meters per second and an angle of 70 degrees. The lower shot has an initial velocity of 31.9 meters per second and an angle of 30 degrees. — **Effigy 4.xvi** Ii trajectories of a golf game ball with a range of 90 thousand. The impact points of both are at the same level as the launch indicate.

Significance

The initial speed for the shot at

$70\text{°}$

is greater than the initial speed of the shot at

$30\text{°}.$

Notation from (Figure) that two projectiles launched at the same speed merely at different angles have the same range if the launch angles add to

$90\text{°}.$

The launch angles in this example add to give a number greater than

$90\text{°}.$

Thus, the shot at

$70\text{°}$

has to have a greater launch speed to accomplish 90 m, otherwise it would country at a shorter distance.

Check Your Agreement

If the two golf shots in (Effigy) were launched at the same speed, which shot would have the greatest range?

[reveal-answer q="fs-id1165166636799″]Show Solution[/reveal-answer]

[subconscious-answer a="fs-id1165166636799″]

The golf game shot at

$30\text{°}.$

[/hidden-answer]

When we speak of the range of a projectile on level ground, we assume R is very small compared with the circumference of Earth. If, however, the range is large, Earth curves away below the projectile and the dispatch resulting from gravity changes direction along the path. The range is larger than predicted by the range equation given before because the projectile has farther to fall than information technology would on level basis, every bit shown in (Figure), which is based on a drawing in Newton's Principia . If the initial speed is great enough, the projectile goes into orbit. Globe's surface drops 5 m every 8000 one thousand. In ane s an object falls 5 grand without air resistance. Thus, if an object is given a horizontal velocity of

$8000\,\text{m}\text{/}\text{s}$

(or

$18,000\text{mi}\text{/}\text{hr})$

near Earth's surface, it will go into orbit around the planet considering the surface continuously falls abroad from the object. This is roughly the speed of the Space Shuttle in a low Earth orbit when it was operational, or any satellite in a depression World orbit. These and other aspects of orbital motion, such every bit Earth's rotation, are covered in greater depth in Gravitation.

The figure shows a drawing of the earth with a tall tower at the north pole and a horizontal arrow labeled v 0 pointing to the right. 5 trajectories that start at the top of the tower are shown. The first reaches the earth near the tower. The second reaches the earth farther from the tower, and the third even farther. The fourth trajectory hits the earth at the equator, and is tangent to the surface at the equator. The fifth trajectory is a circle concentric with the earth. — **Effigy 4.17** Projectile to satellite. In each case shown here, a projectile is launched from a very high tower to avoid air resistance. With increasing initial speed, the range increases and becomes longer than it would be on level basis because Earth curves abroad beneath its path. With a speed of 8000 m/s, orbit is achieved.

Summary

Projectile motion is the motion of an object subject only to the acceleration of gravity, where the acceleration is constant, as about the surface of World.
To solve projectile motility problems, we clarify the motion of the projectile in the horizontal and vertical directions using the i-dimensional kinematic equations for x and y.
The time of flight of a projectile launched with initial vertical velocity
${v}_{0y}$

on an fifty-fifty surface is given by

${T}_{tof}=\frac{2({v}_{0}\text{sin}\,\theta )}{g}.$

This equation is valid just when the projectile lands at the same tiptop from which it was launched.
The maximum horizontal distance traveled by a projectile is called the range. Again, the equation for range is valid merely when the projectile lands at the same elevation from which it was launched.

Conceptual Questions

Answer the following questions for projectile motion on level footing assuming negligible air resistance, with the initial angle being neither

$0\text{°}$

nor

$90\text{°}:$

(a) Is the velocity always zero? (b) When is the velocity a minimum? A maximum? (c) Can the velocity ever be the same as the initial velocity at a time other than at t = 0? (d) Can the speed ever be the aforementioned as the initial speed at a time other than at t = 0?

[reveal-answer q="fs-id1165167780957″]Show Solution[/reveal-answer]

[subconscious-answer a="fs-id1165167780957″]

a. no; b. minimum at apex of trajectory and maximum at launch and impact; c. no, velocity is a vector; d. yes, where information technology lands

[/hidden-answer]

Answer the post-obit questions for projectile motility on level ground assuming negligible air resistance, with the initial angle being neither

$0\text{°}$

nor

$90\text{°}:$

(a) Is the acceleration ever zero? (b) Is the acceleration ever in the aforementioned management as a component of velocity? (c) Is the acceleration ever contrary in direction to a component of velocity?

A dime is placed at the edge of a table so information technology hangs over slightly. A quarter is slid horizontally on the table surface perpendicular to the edge and hits the dime head on. Which money hits the footing showtime?

[reveal-answer q="fs-id1165166623383″]Show Solution[/reveal-answer]

[hidden-answer a="fs-id1165166623383″]

They both hit the footing at the same fourth dimension.

[/hidden-answer]

Problems

A bullet is shot horizontally from shoulder height (1.5 m) with and initial speed 200 m/s. (a) How much time elapses before the bullet hits the ground? (b) How far does the bullet travel horizontally?

[reveal-answer q="fs-id1165168072758″]Evidence Solution[/reveal-answer]

[subconscious-answer a="fs-id1165168072758″]

$t=0.55\,\text{s}$

, b.

$x=110\,\text{m}$

[/hidden-answer]

A marble rolls off a tabletop 1.0 m loftier and hits the flooring at a indicate three.0 chiliad away from the table's border in the horizontal direction. (a) How long is the marble in the air? (b) What is the speed of the marble when it leaves the table's edge? (c) What is its speed when information technology hits the flooring?

A dart is thrown horizontally at a speed of 10 m/s at the balderdash'due south-middle of a dartboard 2.4 m away, equally in the following figure. (a) How far beneath the intended target does the dart hitting? (b) What does your answer tell y'all well-nigh how proficient dart players throw their darts?

[reveal-answer q="fs-id1165168078466″]Show Solution[/reveal-answer]

[subconscious-answer a="fs-id1165168078466″]

$t=0.24\text{s,}\enspaced=0.28\,\text{m}$

, b. They aim loftier.
An illustration of a person throwing a dart. The dart is released horizontally a distance of 2.4 meters from the dart board, level with the bulls eye of the dart board, with a speed of 10 meters per second.

[/subconscious-respond]

An airplane flying horizontally with a speed of 500 km/h at a tiptop of 800 m drops a crate of supplies (see the following effigy). If the parachute fails to open, how far in forepart of the release point does the crate hit the ground?

An airplane releases a package. The airplane has a horizontal velocity of 500 kilometers per hour. The package's trajectory is the right half of a downward-opening parabola, initially horizontal at the airplane and curving down until it hits the ground.

Suppose the aeroplane in the preceding problem fires a projectile horizontally in its direction of move at a speed of 300 1000/s relative to the aeroplane. (a) How far in front of the release point does the projectile hit the footing? (b) What is its speed when it hits the ground?

[reveal-respond q="fs-id1165167989106″]Bear witness Solution[/reveal-respond]

[hidden-answer a="fs-id1165167989106″]

a.,

$t=12.8\,\text{s,}\enspacex=5619\,\text{m}$

${v}_{y}=125.0\,\text{m}\text{/}\text{s,}\enspace{v}_{x}=439.0\,\text{m}\text{/}\text{s,}\enspace|\overset{\to }{v}|=456.0\,\text{m}\text{/}\text{s}$

[/subconscious-respond]

A fastball pitcher tin can throw a baseball at a speed of twoscore m/southward (90 mi/h). (a) Assuming the bullpen tin release the ball 16.7 chiliad from abode plate and so the ball is moving horizontally, how long does it take the ball to reach home plate? (b) How far does the ball drop between the bullpen's hand and abode plate?

A projectile is launched at an angle of

$30\text{°}$

and lands 20 s subsequently at the same height every bit it was launched. (a) What is the initial speed of the projectile? (b) What is the maximum altitude? (c) What is the range? (d) Calculate the deportation from the point of launch to the position on its trajectory at 15 s.

[reveal-answer q="fs-id1165166793284″]Show Solution[/reveal-respond]

[hidden-respond a="fs-id1165166793284″]

${v}_{y}={v}_{0y}-gt,\enspacet=10\text{s,}\enspace{v}_{y}=0,\enspace{v}_{0y}=98.0\,\text{m}\text{/}\text{s},\enspace{v}_{0}=196.0\,\text{m}\text{/}\text{s}$

, b.

$h=490.0\,\text{m},$

${v}_{0x}=169.7\,\text{m}\text{/}\text{s,}\enspacex=3394.0\,\text{m,}$

$\begin{array}{cc} x=2545.5\,\text{m}\hfill \\ y=465.5\,\text{m}\hfill \\ \overset{\to }{s}=2545.5\,\text{m}\hat{i}+465.5\,\text{m}\hat{j}\hfill \end{array}$

[/hidden-answer]

A basketball player shoots toward a basket six.1 m away and three.0 m above the floor. If the ball is released ane.8 grand to a higher place the floor at an angle of

$60\text{°}$

to a higher place the horizontal, what must the initial speed be if it were to go through the basket?

At a particular instant, a hot air balloon is 100 m in the air and descending at a constant speed of ii.0 grand/s. At this exact instant, a girl throws a ball horizontally, relative to herself, with an initial speed of 20 m/s. When she lands, where will she find the brawl? Ignore air resistance.

[reveal-answer q="fs-id1165166591102″]Show Solution[/reveal-answer]

[subconscious-reply a="fs-id1165166591102″]

$-100\,\text{m}=(-2.0\,\text{m}\text{/}\text{s})t-(4.9\,\text{m}\text{/}{\text{s}}^{2}){t}^{2},$

$t=4.3\,\text{s,}$

$x=86.0\,\text{m}$

[/hidden-answer]

A man on a motorcycle traveling at a uniform speed of 10 grand/s throws an empty can straight upward relative to himself with an initial speed of iii.0 m/s. Find the equation of the trajectory equally seen by a police officer on the side of the road. Presume the initial position of the can is the point where it is thrown. Ignore air resistance.

An athlete can bound a distance of 8.0 m in the wide bound. What is the maximum altitude the athlete can jump on the Moon, where the gravitational acceleration is one-sixth that of Earth?

[reveal-respond q="fs-id1165167996165″]Testify Solution[/reveal-answer]

[hidden-answer a="fs-id1165167996165″]

${R}_{Moon}=48\,\text{m}$

[/hidden-answer]

The maximum horizontal distance a male child can throw a ball is 50 grand. Assume he can throw with the same initial speed at all angles. How high does he throw the brawl when he throws it straight up?

A stone is thrown off a cliff at an angle of

$53\text{°}$

with respect to the horizontal. The cliff is 100 m high. The initial speed of the rock is thirty grand/due south. (a) How high above the border of the cliff does the rock ascent? (b) How far has information technology moved horizontally when it is at maximum distance? (c) How long later on the release does it striking the ground? (d) What is the range of the rock? (eastward) What are the horizontal and vertical positions of the rock relative to the edge of the cliff at t = 2.0 southward, t = 4.0 s, and t = half-dozen.0 southward?

[reveal-answer q="fs-id1165167746378″]Testify Solution[/reveal-answer]

[hidden-respond a="fs-id1165167746378″]

$\[\]$

${v}_{0y}=24\,\text{m}\text{/}\text{s}$

${v}_{y}^{2}={v}_{0y}^{2}-2gy⇒h=23.4\,\text{m}$

,
b.

$t=3\,\text{s}\enspace{v}_{0x}=18\,\text{m/s}\enspacex=54\,\text{m}$

$y=-100\,\text{m}\enspace{y}_{0}=0$

$y-{y}_{0}={v}_{0y}t-\frac{1}{2}g{t}^{2}\enspace-100=24t-4.9{t}^{2}$

$⇒t=7.58\,\text{s}$

$x=136.44\,\text{m}$

$t=2.0\,\text{s}\enspacey=28.4\,\text{m}\enspacex=36\,\text{m}$

$t=4.0\,\text{s}\enspacey=17.6\,\text{m}\enspacex=22.4\,\text{m}$

$t=6.0\,\text{s}\enspacey=-32.4\,\text{m}\enspacex=108\,\text{m}$

[/hidden-answer]

Trying to escape his pursuers, a secret agent skis off a gradient inclined at

$30\text{°}$

below the horizontal at 60 km/h. To survive and country on the snow 100 m below, he must articulate a gorge 60 m wide. Does he make information technology? Ignore air resistance.
A skier is moving with velocity v sub 0 down a slope that is inclined at 30 degrees to the horizontal. The skier is at the edge of a 60 m wide gap. The other side of the gap is 100 m lower.

A golfer on a fairway is 70 m away from the light-green, which sits below the level of the fairway by 20 k. If the golfer hits the ball at an bending of

$40\text{°}$

with an initial speed of 20 grand/due south, how shut to the green does she come up?

[reveal-respond q="fs-id1165168065281″]Show Solution[/reveal-respond]

[hidden-respond a="fs-id1165168065281″]

${v}_{0y}=12.9\,\text{m}\text{/}\text{s}\,y-{y}_{0}={v}_{0y}t-\frac{1}{2}g{t}^{2}\enspace-20.0=12.9t-4.9{t}^{2}$

$t=3.7\,\text{s}\enspace{v}_{0x}=15.3\,\text{m}\text{/}\text{s}⇒x=56.7\,\text{m}$

And then the golfer's shot lands 13.3 m short of the light-green.

[/subconscious-answer]

A projectile is shot at a hill, the base of operations of which is 300 g away. The projectile is shot at

$60\text{°}$

to a higher place the horizontal with an initial speed of 75 m/s. The hill tin be approximated past a plane sloped at

$20\text{°}$

to the horizontal. Relative to the coordinate system shown in the post-obit figure, the equation of this straight line is

$y=(\text{tan}20\text{°})x-109.$

Where on the loma does the projectile country?
A projectile is shot from the origin at a hill, the base of which is 300 m away. The projectile is shot at 60 degrees above the horizontal with an initial speed of 75 m/s. The hill is sloped away from the origin at 20 degrees to the horizontal. The slope is expressed as the equation y equals (tan of 20 degrees) times x minus 109.

An astronaut on Mars kicks a soccer ball at an angle of

$45\text{°}$

with an initial velocity of xv m/s. If the acceleration of gravity on Mars is three.seven thousand/s, (a) what is the range of the soccer kick on a flat surface? (b) What would be the range of the same boot on the Moon, where gravity is one-6th that of World?

[reveal-answer q="fs-id1165166572087″]Show Solution[/reveal-reply]

[hidden-answer a="fs-id1165166572087″]

$R=60.8\,\text{m}$

,
b.

$R=137.8\,\text{m}$

[/hidden-answer]

Mike Powell holds the record for the long jump of 8.95 m, established in 1991. If he left the ground at an bending of

$15\text{°},$

what was his initial speed?

MIT's robot chetah can jump over obstacles 46 cm high and has speed of 12.0 km/h. (a) If the robot launches itself at an bending of

$60\text{°}$

at this speed, what is its maximum superlative? (b) What would the launch angle take to be to reach a pinnacle of 46 cm?

[reveal-answer q="fs-id1165167842253″]Bear witness Solution[/reveal-answer]

[hidden-answer a="fs-id1165167842253″]

${v}_{y}^{2}={v}_{0y}^{2}-2gy⇒y=2.9\,\text{m}\text{/}\text{s}$

$y=3.3\,\text{m}\text{/}\text{s}$

$y=\frac{{v}_{0y}^{2}}{2g}=\frac{{({v}_{0}\text{sin}\,\theta )}^{2}}{2g}⇒\text{sin}\,\theta =0.91⇒\theta =65.5\text{°}$

[/hidden-respond]

Mt. Asama, Nihon, is an active volcano. In 2009, an eruption threw solid volcanic rocks that landed i km horizontally from the crater. If the volcanic rocks were launched at an bending of

$40\text{°}$

with respect to the horizontal and landed 900 m below the crater, (a) what would be their initial velocity and (b) what is their time of flight?

Drew Brees of the New Orleans Saints can throw a football game 23.0 m/s (50 mph). If he angles the throw at

$10\text{°}$

from the horizontal, what distance does it get if information technology is to be caught at the aforementioned peak as it was thrown?

[reveal-answer q="fs-id1165168098591″]Bear witness Solution[/reveal-answer]

[hidden-reply a="fs-id1165168098591″]

$R=18.5\,\text{m}$

[/hidden-answer]

The Lunar Roving Vehicle used in NASA's late Apollo missions reached an unofficial lunar land speed of 5.0 m/s past astronaut Eugene Cernan. If the rover was moving at this speed on a apartment lunar surface and hit a small bump that projected it off the surface at an angle of

$20\text{°},$

how long would it be "airborne" on the Moon?

A soccer goal is ii.44 grand loftier. A player kicks the brawl at a altitude 10 1000 from the goal at an angle of

$25\text{°}.$

What is the initial speed of the soccer ball?

[reveal-answer q="fs-id1165167854326″]Show Solution[/reveal-answer]

[hidden-reply a="fs-id1165167854326″]

$y=(\text{tan}\,{\theta }_{0})x-[\frac{g}{2{({v}_{0}\text{cos}\,{\theta }_{0})}^{2}}]{x}^{2}⇒{v}_{0}=16.4\,\text{m}\text{/}\text{s}$

[/hidden-answer]

Olympus Mons on Mars is the largest volcano in the solar system, at a height of 25 km and with a radius of 312 km. If you are standing on the summit, with what initial velocity would you have to burn down a projectile from a cannon horizontally to clear the volcano and land on the surface of Mars? Notation that Mars has an dispatch of gravity of

$3.7\,\text{m}\text{/}{\text{s}}^{2}.$

In 1999, Robbie Knievel was the first to jump the Grand Canyon on a motorcycle. At a narrow part of the canyon (69.0 m wide) and traveling 35.8 thou/s off the takeoff ramp, he reached the other side. What was his launch angle?

[reveal-answer q="fs-id1165168009639″]Show Solution[/reveal-answer]

[hidden-reply a="fs-id1165168009639″]

$R=\frac{{v}_{0}^{2}\text{sin}\,2{\theta }_{0}}{g}⇒{\theta }_{0}=15.0\text{°}$

[/hidden-answer]

You throw a baseball at an initial speed of fifteen.0 m/s at an angle of

$30\text{°}$

with respect to the horizontal. What would the ball'due south initial speed take to be at

$30\text{°}$

on a planet that has twice the dispatch of gravity as Earth to attain the same range? Consider launch and impact on a horizontal surface.

Aaron Rogers throws a football at 20.0 chiliad/s to his wide receiver, who runs straight downwards the field at 9.iv m/s for xx.0 m. If Aaron throws the football when the wide receiver has reached 10.0 grand, what angle does Aaron take to launch the ball so the receiver catches it at the 20.0 one thousand mark?

[reveal-answer q="fs-id1165166777489″]Show Solution[/reveal-answer]

[subconscious-answer a="fs-id1165166777489″]

It takes the wide receiver 1.i s to cover the last 10 chiliad of his run.

${T}_{\text{tof}}=\frac{2({v}_{0}\text{sin}\,\theta )}{g}⇒\text{sin}\,\theta =0.27⇒\theta =15.6\text{°}$

[/subconscious-answer]

Glossary

projectile motion: move of an object discipline only to the acceleration of gravity

range: maximum horizontal altitude a projectile travels

time of flying: elapsed fourth dimension a projectile is in the air

trajectory: path of a projectile through the air

tookesackled.blogspot.com

Source: https://opentextbc.ca/universityphysicsv1openstax/chapter/4-3-projectile-motion/

A Projectile Is Thrown Directly Upward and Caught Again. At the Top of Its Path

four.3 Projectile Move

Learning Objectives

Problem-Solving Strategy: Projectile Motion

Instance

A Fireworks Projectile Explodes High and Away

Strategy

Solution

Check Your Understanding

Example

Calculating Projectile Move: Tennis Actor

Strategy

Solution

Significance

Time of Flight, Trajectory, and Range

Time of flying

Trajectory

Range

Example

Comparing Golf game Shots

Strategy

Solution

Significance

Check Your Agreement

Summary

Conceptual Questions

Problems

Glossary

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