Asked by Kalyn

A diver of mass 43.2 kg jumps off a high dive 10.0 m tall. If he comes to a stop 2.0 m below the surface of the water, what was his acceleration after he entered the water?

Answers

Answered by drwls
Vo = sqrt(2 a X)
Solve for acceleration, a.
X is the distance he goes below the water surface.
Vo is his speed when he hits the water.
Vo = sqrt(2 g H)

2 a X = 2 g H

a = g H / X
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