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 34.5 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 0.710 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 123 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?
after time t = 0.71 the ball is level with Bob
the vertical component of velocity = voy
displacement dy = 0
ay = -g = -9.8 m/s^2
from equations of motion
dy = voy*t + 0.5*g*t^2
0 = voy*0.71 - 0.5*9.8*0.71^2
voy = 3.48 m/s
vo = 34.5
vo^2 = vox^2 + voy^2
horizontal component of velocity = vox = sqrt(34.5^2-3.48^2) = 34.3 m/s
vox = 34.3 m/s
range = R = 123 = vox*T
T = 123/34.3
T = 3.6 s
for the total motion of the ball
displacement = -(h+2)
time = T = 3.6 s
voy = 3.48 m/s
-(h+2) = 3.48*3.6 - 0.5*9.8*3.6^2
h = 48.976 m <<---------answer
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