a rock is dropped from a height of 15m above the ground. when it hits the ground, it leaves a hole 12cm deep. what is the speed of the rock when it hits the ground? what is the deceleration of the rock when it hits the ground?

Answer 1

1. Vf^2 = Vo^2 + 2g*d,
Vf^2 = 0 + 19.6*15 = 294,
Vf = 17.15m/s.

2. a = (Vf^2 - Vo^2) / 2d,
a = (0 - (17.15)^2) / 24 = -12.26s.

Answer 2

Correction: a = -12.26m/s^2.