What is the momentum of a 5 kg ball dropped from a 45 m high window the instant before it hits the ground, in kg·m/s?

Ans: 150 kg*m/s

mgh=(1/2)mv^2
v^2=2gh
=2(9.8 m/s^2)(45 m)
=882
v=29.698 m/s^2
so, p=mv
=(5 kg)(29.698 m/s^2) = 148.6 kg*m/s = 150 kg*m/s (sig figs)

Physics & Space Science

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