Question

A 0.5 kg wrench is dropped from a height of 10 meters. Using v = square root of2gh, where g = 9.80 m/s^2, what is the momentum just before it strikes the floor?

Answer 1

$mgh= \frac{1}{2}mv^2 \\ v= \sqrt{2gh} \\ = \sqrt{2 \times 9.8 \times 10} \\ = 14.0 \;m/s \\ p =mv \\ = 0.5 \times 14 \\ = 7.0 \;kg \;m/s, \;dowm$