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 34.1 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 124 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,

maximum initial speed , u = 32.9 m/s

time of flight , t = 0.910 s

let the angle of projection is theta

for the time of same level

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

0.910 = 2 * 32.9 * sin(theta)/9.8

solving for theta

theta = 7.82 degree

Here, let the height of cliff is h

time of flight is tf

tf = distance/(v * cos(theta))

tf = 128/(32.9 * cos(7.82 degree))

tf = 3.93 s

Using second equation of motion

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

- (h + 2) = 32.9 * sin(7.89) * 3.93 - 0.5 * 9.8 * 3.93^2

solving

h = 55.9 m

the height of the cliff is 55.9 m

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