A 1.0 kg ball falls from rest a distance of 19.6 m. - What was its change in potential energy?
- What was its velocity after falling 19.6 m?

192.08J

19.6m/s

Explanation:

Since there will be no potential energy when the ball is on the ground, the change in potential energy is equal to the potential energy at the start when the ball is 19.6m above the ground.

PE=mgh

=(1)(9.8)(19.6)

=192.08J

v²=u²+2as, where v is the final velocity, u is initial velocity, a is acceleration and s is distance. Initial velocity is 0 since it starts at rest.

v²=u²+2as

v²=0²+2(9.8)(19.6)

v=√384.16

=19.6m/s

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