Bob has just finished climbing a sheer cliff above a beach a

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 76.0 mph. 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 0.710 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 354 ft from the base of the cliff. How high up is Bob, if the ball started from exactly 5 ft above the edge of the cliff?

Solution

time to return to Bob level is t=2v_0y/g
then v_0y=tg/2
v_0y=0.710*9.8/2=3.48 m/s
v₀sinx=3.48
x=sin^-1(3.48/32.5)=6.15 degree
v_0x=32.5 m/s
then
x=v_0xt and t=x/v_0x=3.94 s (fly time)
h_max=v_0y^2/(2g)=0.62 m (respect starting point)
h_max is reached in t=v_oy/g=0.355=0.710/2
from this point the eq of motion in vertical is
y=1/2gt^2 (t=3.94-0.71/2=3.585 s)
y=62.98 m
h=62.98-2- 0.62=60.36 m