A body A of mass 3 . 0 kg and a body B of mass 10 kg are dropped simultaneously from a height of 14 . 9 m Calculate
their potential energies, and

Explanation

As the two bodies are dropped, they fall with the same acceleration of ` 9 . 8 m s^(-2)` . When they are 10 m above the ground, they have already fallen through 14. 9m = - 10 m = 4 . 9 m The velocity at this point may be worked out form
`v^(2) = u^(2) + 2 gh = 0 + 2 xx (9 . 8 m s^(-2)) xx ( 4 . 9 m)`
`v^(2) = 9 . 8 xx 9 . 8 m^(2) s^(-2) or v = 9 . 8 m s^(-1)`
The bodies are at a height of 10 m above the ground
The potential enegy of body A is
`U_(A) = m_(A) gh`
`= ( 3 . 0 kg) xx (9 . 8 m s^(-2)) xx (10 m) = 294 J`
The potential energy of body B is
`U_(B) = m_(B) gh`
`= (10 kg) xx (9 . 8 m s^(-2)) xx (10 m ) = 980 J`