Suppose you "dig a hole" 52 feet deep and place a baseball machine in the hole, so that the balls fire out at 96 ft /s and land on the ground outside the hole. What is the velocity of one of these baseballs when it hits the ground?

76.73 ft/s

Explanation:

Let the final velocity is v.

initial velocity, u = 96 ft/s

g = 32 ft/s²

height, h = 52 feet

use third equation of motion

v² = u² - 2 gh

v² = 96 x 96 - 2 x 32 x 52

v = 76.73 ft/s

Thus, the speed of the ball as it reaches the ground is 76.73 ft/s.

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