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Posted by Taylor on Wednesday, September 21, 2011 at 4:26pm.

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?

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.

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

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