News: Currently the LaTeX and hidden solutions on this blog do not work on Google Reader.
Email me if you have suggestions on how to improve this blog!

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.


Post a Comment

This webpage is LaTeX enabled. To type in-line formulae, type your stuff between two '$'. To type centred formulae, type '\[' at the beginning of your formula and '\]' at the end.