Question

a ball is dropped from a height h the potential energy at a particular instant is nine times of kinetic energy. When the velocity os double, what will be the ratio of new kinetic and potential energy?

Answer 1

Answer:Hello Vaishnavi dear, as body is dropped from a height h, its energy = m g h

When its comes down through a distance x then its potential energy = m g ( h - x )

So KE attained has to be m g x because of conservation of energy

Given PE = 9 * KE

==> m g ( h - x) = 9 * m g x

So x = h/10

As velocity is to be two times of the velocity attained at this location then new KE would be 4 * old KE

​Hence KE (New) = 4 * m g x = 4 * m g * h/10

OR 2/5 * mg h

If so, then PE = 3/5 m g h (as per conservation of energy)

Hence the required ratio of new KE and new PE = 2 : 3

Explanation: