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  4. Motion in two and three dimensions
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4.3 Projectile motion  (Page 4/13)

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Check Your Understanding A rock is thrown horizontally off a cliff 100.0 m high with a velocity of 15.0 m/s. (a) Define the origin of the coordinate system. (b) Which equation describes the horizontal motion? (c) Which equations describe the vertical motion? (d) What is the rock’s velocity at the point of impact?

(a) Choose the top of the cliff where the rock is thrown from the origin of the coordinate system. Although it is arbitrary, we typically choose 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 . (c) [link] through [link] and [link] describe the vertical motion, but since y 0 = 0 and v 0 y = 0 , these equations simplify greatly to become y = 1 2 ( v 0 y + v y ) t = 1 2 v y t , v y = − g t , y = − 1 2 g t 2 , and v y 2 = −2 g y . (d) We use the kinematic equations to find the x and y components of the velocity at the point of impact. Using v y 2 = −2 g y and noting the point of impact is −100.0 m, we find the y component of the velocity at impact is v y = 44.3 m / s . We are given the x component, v x = 15.0 m / s , so we can calculate the total velocity at impact: v = 46.8 m/s and θ = 71.3 ° below the horizontal.

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Calculating projectile motion: tennis player

A tennis player wins a match at Arthur Ashe stadium and hits a ball into the stands at 30 m/s and at an angle 45 ° above the horizontal ( [link] ). On its way down, the ball is caught by a spectator 10 m above the point where the ball was hit. (a) Calculate the time it takes the tennis ball to reach the spectator. (b) What are the magnitude and direction 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.
The trajectory of a tennis ball hit into the stands.

Strategy

Again, resolving this two-dimensional motion into two independent one-dimensional motions allows us to solve for the desired quantities. The time a projectile is in the air is governed by its vertical motion alone. Thus, we solve for t first. While the ball is rising and falling vertically, the horizontal motion continues at a constant velocity. This example asks for the final velocity. Thus, we recombine the vertical and horizontal results to obtain v → at final time 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 final position 10.0 m higher than its starting altitude. We can find the time for this by using [link] :

y = y 0 + v 0 y t − 1 2 g t 2 .

If we take the initial position y 0 to be zero, then the final position is y = 10 m. The initial vertical velocity is the vertical component of the initial velocity:

v 0 y = v 0 sin θ 0 = ( 30.0 m / s ) sin 45 ° = 21.2 m / s .

Substituting into [link] for y gives us

10.0 m = ( 21.2 m/s ) t − ( 4.90 m/s 2 ) t 2 .

Rearranging terms gives a quadratic equation in t :

( 4.90 m/s 2 ) t 2 − ( 21.2 m/s ) t + 10.0 m = 0 .

Use of the quadratic formula yields t = 3.79 s and t = 0.54 s. Since the ball is at a height of 10 m at two times during its trajectory—once on the way up and once on the way down—we take the longer solution for the time it takes the ball to reach the spectator:

t = 3.79 s .

The time for projectile motion is determined completely by the vertical motion. Thus, any projectile that has an initial vertical velocity of 21.2 m/s and lands 10.0 m below its starting altitude spends 3.79 s in the air.
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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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