Bob is a pitcher, and knows that the fastest he can throw the ball is about 32.1

Question

Bob is a pitcher, and knows that the fastest he can throw the ball is about 32.1 m/s. Bob starts the stopwatch as he throws the ball (with no way to measure the ball's initial trajectory), and watches carefully. The ball rises and then falls, and after 0.510 seconds the ball is once again level with Bob. Bob can't see well enough to time when the ball hits the ground. Bob's friend then measures that the ball landed 127 m from the base of the cliff. How high up is Bob, if the ball started from exactly 2 m above the edge of the cliff?

Explanation / Answer

1) time at/above launch height t = 2·Vo·sin/g

0.510 s = 2 * 32.1 * sin / 9.81 = 6.92s * sin
= arcsin(0.510/6.5443) = 4.47º
Then the horizontal velocity is Vx = 32.1 m/s * cos4.47º = 32.00 m/s

Then the ball was between launch height and the base of the cliff for time

t = x / v = 127 m / 32.00 m/s = 3.968 s

The initial vertical velocity is Vy = 32.1 m/s * sin4.47º = 2.501 m/s
Upon returning to launch height, then, Vv = - 2.501 m/s

Then the vertical displacement from launch height to the base of the cliff is

y = Vv*t + ½at² = -2.501 m/s * 3.968s - ½ * 9.81 m/s² * (3.968s)² = 87.15 m

Therefore the cliff height is 87.15 m - 2 m =85.15 m Ans.

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