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, 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 faster he can throw the ball is about 34.1 m/s. Bob s(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 126 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,

u = 34.1 m/s

let the angle above the horizontal be theta

after 0.71 s

h = 0

u * sin(theta) * t - 0.5 * g * t^2 = 0

34.1 * sin(theta) - 0.5 * 9.8 * t = 0

theta = 5.86 degree

let the total time taken be t1

u * cos(theta) * t1 = 126 m

34.1 * cos(5.86) * t1 = 126

t1 = 3.71 s

let the height of cliff be h

- (h - 2) = u * sin(theta) * t1 - 0.5 * g * t1^2

- (h-2) = 34.1 * sin(5.85) * 3.71 - 0.5 * 9.8 * 3.71^2

solving for h

h = 56.5 m

the height of the cliff is 56.5 m

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