Use the following energy relations [tex]E_{Grav}=mgh[/tex] [tex]E _{Kinetic} = \frac{1}{2}m v^{2} [/tex] Then look at momentum put into the kinetic energy formula. Recall [tex]p=mv[/tex] Now we see that [tex]E _{Kinetic}= \frac{ p^{2} }{2m} [/tex] Since the kinetic energy at the end of the fall is the same as the potential (Gravitational) energy at the beginning we set them equal [tex]mgh= \frac{ p^{2} }{2m} [/tex] [tex] m\sqrt{2gh} = p=62.61 \frac{kgm}{s} [/tex]
