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 33.7 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 0.910 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

Here,

let theta be angle of flight

as time to reach the same height , tf = 0.910

tf = 2*(v*sin(theta))/g

0.910 = 2 * (33.7 * sin(theta))/9.8

sin(theta) = 0.132

theta = 7.6 degree

Now, time of flight , t

t = 127/(33.7 * cos(theta))

t = 127/(33.7 * cos(7.6))

t = 3.802 s

Now, using second equation of motion for finding the height ,

- (h + 2) = 33.7 * sin(7.6) * 3.802 - 0.5 * 9.8 * 3.802^2

solving for h

h = 51.88 m

the height of the cliff is 51.88 m

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