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 33.7 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 127 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?

Answer 1

Velocity = 33.7 m/s
We can consider (v(x)^2 +v(y)^2)^1/2 = 33.7
time up = time down ===> time up = 0.71 s / 2 = 0.355 s

v(y) = at = 9.80m/s^2 * 0.355 s = 3.479 m/s Vertical component of velocity

So we know the hypotenuse and the opposite side===> We can find the Launch Angle

Sin(a) = 3.479 / 33.7 = 5.925°

a = 5.925° ===> Now we can find the horizontal component of velocity

cos5.925° = x/33.7

v(x) = 33.52 m/s

How high did the ball go?

d = 0.5at^2
d = 0.5 * 9.8 m/s^2 * 0.355^2
d = 0.617 m The ball fell from there under gravity

so the total time of flight is
t = d/v(x)
t = 127 /33.52
t = 3.788 s

Subtract the time up to get the time to fall

3.788 s - 0.355 s = 3.433 s to fall from maximum height

d = 0.5at^2
d = 0.5 * 9.8 m/s * 3.433^2
d = 57.748 m

Subtract the height from Bob's hand to the max height

57.748 - 0.617 = 57.131 m

Subtract Bob's height

57.131 - 5 = 52.131 m height of the cliff