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]
