Question

A 131 kg car is at rest at the top of a 39 m high hill. If the car rolls down a frictionless hill, how fast will it be moving when it reaches ground level?

Answer 1

27,6 m/s

Explanation:

The potential energy of the car (mgh) will become kinetic energy at the ground level (`1/2 mv^2`)
`m xx g xx h = 1/2 xx m xx v^2`
if you neglet the friction of ground and of air it not important the mass
`g xx h = 1/2 xx v^2`
so, you can obtain the velocity
`v = sqrt( 2g xx h) = sqrt (2 xx 9,8 m/s^2 xx 39 m) =27,6 m/s`