A ball is dropped from a height 64 m above the ground and every time it hits the ground it rises to a height equal to half of the previous height. The height attained after it hits the ground for the 16h time is

A ball is dropped from a height 64 m above the ground and every time it hits the ground it rises to a height equal to half of the previous. What is the height attained after it hits the ground for the 16^(th) time?

A ball dropped from a height 'h' reaches the ground in time 'T'. What is its height from the ground at time T/2 ?

A ball is dropped from a height h above ground. Neglect the air resistance, its velocity (v) varies with its height (y) above the ground as :-

A ball is dropped from a height h above ground. Neglect the air resistance, its velocity (v) varies with its height (y) above the ground as :-

A ball is dropped from a height h above ground. Neglect the air resistance, its velocity (v) varies with its height (y) above the ground as :-

An object is dropped from a height h from the ground .Every time it hits the ground it looses 50% of its kinetic energy.The total distance covered as t to oo is:

A ball is dropped from a height h above ground,Neglect the air resistance,its velocity (v) varies with its height y above the ground as

An object is dropped from a height h from the ground. Every time it hits the ground it looses 50% of its kinetic energy. The total distance covered as t to oo is -

A ball is thrown upwards with a speed u from a height h above the ground.The time taken by the ball to hit the ground is