Question

Does anyone know how to do this problem

Bob has just finished climbing a sheer cliff above a beach, and wants to figure out how high ne 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 81.0 mph- Bob starts the stopwatch as he throws toe bal (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 387 ft from the base of the diff. How high up is Bob, if the ball started from exactly 5 ft above the edge of toe cliff?

Answer 1

81 mph = 118.8 ft/s

time of flight = 2vsintheta/g

0.910 = 2*118.8 * sin theta /32

theta = 7.04 degree

Vx = V cos theta = 117.9 ft/s

Vy = 14.56 ft/s

since the range is 387 ft

t = R/Vx = 3.28s

total time from cliff to ground = 3.28 - 0.910 = 2.37 s

s = vy*t + 1/2*gt^2

s = 124.38 ft