A ball is thrown straight up with an initial speed of 20 m/s. The ball was released 2 m above the ground, but when it returns back down, it falls into a hole 11.5 m deep. What is the ball’s speed (in m/s) at the bottom of the hole? You may treat this as an isolated system.

Explanation:

height, h = 20 m

initial speed, u = 20 m/s

depth of hole, d = 11.5 m

Let the speed of the ball as it strikes the ground is v.

use third equation of motion

v² = u² + 2gh

v² = 0 + 2 x 9.8 x 2

v² = 39.2 m/s

Let it strikes teh bottom of hole with speed v'

Use third equation of motion

v'² = v² + 2 gd

v'² = 39.2 + 2 x 9.8 x 11.5

v'² = 264.6

v' = 16.3 m/s

answer; how far out of place which the object finished relative to the starting location;