Bob has just finished climbing a sheer cliff above a beach, and wants to figure
ID: 1443150 • Letter: B
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 32.9 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 0.910 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 129 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
v = 32.9 m/s t = 0.91 s, d= 129 m, 2m
From kinematic equation along verticla direction
hf =hi +vi*t -(1/2) gt^2
hf = hi
vi = 0.5*9*t
vi =0.5*9.8*0.91 = 4.459 m/s
vi = vsin(theta)
theta = arcsin(4.459/32.9) = 7.79 degrees
horizontal component vx = vcos(theta)
vx = 32.9cos(7.79) =32.6 m/s
t = distance/velocity
t = 129/ 32.6
t = 3.96 s
from kinematic equation along vertical direction
hf =0
hi = -vi*t+(1/2)gt^2
hi = -(4.459*3.96) + (0.5*9.8*3.96*3.96)
hi =59.18 m
finally the ball started 2m above the cliff, so
cliff height =hi -2 = 59.18 -2
= 57.18 m