Question
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?
Answers
Henry
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.
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.
Henry
Correction: a = -12.26m/s^2.