A 5kg ball is dropped from a height of 10 m. (a) Find the initial potential energy of the ball. (b) Find the kinetic energy, just before it reaches the ground and (c ) Calculate the velocity before it reaches the ground.

Explanation

Here mass of the ball, `m =5kg`
height of the ball, `h= 10m`
(a) initial potential energy of the ball,
`E_p = mgh = (5kg)(10m//s^2)(10m) = 500J`
(b) When the ball reaches the ground, its potential energy becomes zero as it is entirely converted into
its kinetic energy `(E_k),i.e.,`
`E_k =500J ...(i)`
(c ) It `upsilon` is the velocity attained by the ball before reaching the ground,
`E_k =(1)/(2)m upsilon^2`
or `upsilon =sqrt((2E_k)/(m)) = sqrt((2xx500J)/(5kg))= 14.14m//s`