The ball is projected vertically down with a velocity of 10sqrt(3)m/s from a height of 20m .On the hitting the ground if it loses 40% of its energy the height to which it bounces is (g=10ms^(-2))

A ball is thrown vertically downward with a velocity of 20m/s from the top of a tower,It hits the ground after some time with a velocity of 80m/s .The height of the tower is : (g=10m/g^(2))

A ball is projected vertically down with an initial velocity from a height of 20 m onto a horizontal floor. During the impact it loses 50% of its energy and rebounds to the same height. The initial velocity of its projection is

A ball is projected vertically up with speed 20 m/s. Take g=10m//s^(2)

A ball initally at rest is dropped from a height of 10 m . In strinking the ground , it loses 20 % of its kinetic energy . Calculate the height to which it bounces . Where does the lost kinetic energy go ?

A ball falls from a height of 10 m. On rebounding, it loses 30% energy. The ball goes upto a height of

A ball of mass 2 kg is projected horizontally with a velocity 20m/s from a building of height 15m. The speed with which body hits the ground is :

A ball at rest is dropped freely from a height of 20 m . It loses 30 % of its energy on striking the ground and bounces back. The height to which it bounces back is

A ball is dropped from rest from a height of 0m. If it loses 20% of its kinetic energy on striing the ground, what is the height to which the ball bounces?

A ball is projected vertically up wards with a velocity of 100 m/s. Find the speed of the ball at half the maximum height. (g=10 m//s^(2))