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Saturday, 13 August 2011

Physics GRE - #12

A ball is dropped from a height h. As it bounces off the floor, its speed is 80% of what it was just before it hit the floor. The ball will then rise to a height of about:

0.94 h

0.80 h

0.75 h

0.64 h

0.50 h

Solution :

Conservation of energy tells us that $mgh = \frac{1}{2}mv^2$. Therefore, we know that $h\propto v^2$. Since $v_{new}=0.8v_{old}$, $h_{new} = (0.8)^2h_{old} = 0.64h$.

It's a simple problem, but it may just mess you up if you aren't careful.

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