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 32.5 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 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?

please be extremely specific when showing steps!!!!!

Explanation / Answer

In the vertical direction, we can say the ball went to a height in half the time given to go up and down, and it will momentarily stop at the height under the acceleration of gravity

Therefore apply vf = vo + at

0 = vo + (9.8)(.255)

vo = 2.499 m/s

That is the y velocity initial

We can then apply vy = sin(angle)(v)

2.499 = 32.5(sin angle)

angle = 4.41 degrees

The x component of the velocity will be 32.5(cos 4.41)

vx = 32.4 m/s

Then d = vt to find the time to travel the horizontal distance

126 = 32.4(t)

t = 3.888 sec

That time is used to travel downward to the ground

d = vot + .5at2

d = (2.499)(3.88) + (.5)(-9.8)(3.88)2

d = -64.4 m

Then Bob is 64.4 - 2 m high

That is 62.4 m

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