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A freely falling body strikes a horizontal surface and raises to a height $$\frac{1}{64}$$ of the height from which it was dropped. If the velocity with which it hits the floor is $$v$$, the change in the magnitude of velocity due to impact with the floor is
If the body is dropped from height $$h$$, then $$v^2 = 2gh$$, and the rebound speed $$u$$ satisfies $$u^2 = 2g \cdot \frac{h}{64}$$, so $$u = \frac{v}{8}$$. The change in the magnitude of velocity (speed) is $$v - \frac{v}{8} = \frac{7v}{8}$$.
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