Bob has just finished climbing a sheer cliff above a beach, and wants to figure

Question

Bob has just finished climbing a sheer cliff above a beach, and wants to figure out how high he climbed. All he has to use, however, is a baseball, a stopwatch, and a friend on the ground below with a long measuring tape. 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 0.710 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 121m from the base of the cliff. How high up is Bob, if the ball started from exactly 2m above the edge of the cliff?

Explanation / Answer

Here ,

initial speed , v0 = 32.1 m/s

as time to reach same height , t = 0.710 s

let the angle of projection is theta

for time of flight

t = 2 * v0 * sin(theta)/g

0.710 = 2 * 32.1 * sin(theta)/9.8

sin(theta) = 0.109

theta = 6.22 degree

Now, as the horizontal distance, R = 121 m

let the time of flight is tf

tf = 121/(32.1 * cos(6.22))

tf = 3.79 s

Now, let height is h

using second equation of motion

- (h + 2) = 32.1 * sin(6.22)* 3.79 - 0.5 * 9.8 * 3.79^2

h = 55.2 m

the height of the cliff is 55.2 m

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