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 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 32.5m/s. bob starts the stopwatch as he throws the ball and watches carefully. the ball rises and then falls, and after 0.510 seconds the ball is once again level with bob. bob's friend then measures that the ball landed 123m from the base of the cliff. how high is bob, if the ball started from 2m?

In order to solve this problem, you need to divide the motion of the ball into its vertical and horizontal components.

Consider the vertical component: the time taken to rise and fall at Bob's level again is twice the time taken to reach the highest point. At the highest point, the speed (v₁) is zero. Therefore, we can calculate the vertical component of the initial velocity considering it is a decelerated motion:
v₁ = vy - g·t
vy = g·t
    = (9.8)·(0.255)
    = 2.5 m/s

Since you know the magnitude of the initial speed and now also the vertical component, you can calculate the angle of the direction:
vy = v₀ · sinα
α = sin⁻¹ (vy / v₀)
   = sin⁻¹ (0.0769)
   = 4.4°

Now, we can calculate the horizontal component of the initial velocity:
vx = v₀ · cosα
    = 32.5 · cos(4.4)
    = 32.4 m/s

Horizontally, the ball is moving at a constant speed, therefore we can find the total time traveled:
d = vx · t
t = d / vx
  = 123 / 32.4
  = 3.796 s

This is the time taken for the ball to reach the ground from when Bob launches it. Subtracting the time taken to rise and fall again at Bob's level, you get the time taken to reach the ground from Bob's level:
t₂ = t - t₁
    = 3.796 - 0.510
    = 3.286 s

This last part of the motion is a free fall from a position h₀ = 2 meters above the cliff (which is Bob's height), with initial speed vy = 2.5 m/s during a time t₂ = 3.286 s; you can then solve for the height of the cliff:
h = h₀ + vy · t₂ + (1/2) · g · t₂²
   = -2 + (2.5)(3.286) + (0.5)(9.8)(3.286)²
   = -2 + 8.215 + 52.909
   = 60.124 m

Therefore, the correct answer is: the cliff is 60m high.

The objects do not reach terminal velocity         there is not enough time from the time the object falls from when it hits the ground to achieve terminal velocity. an object must be placed higher to achieve terminal velocity in order to be in free fall.