A ball is released from the top of a tower of height h metre. It takes T second to reach the ground. What is the position of the ball in `T/3` second?

A body is released from the top of a tower of height h. It takes t sec to reach the ground. Where will be the ball after time (t)/(2) sec?

A ball is relased from the top of a building 180m high. It takes time t to reach the ground With what speed should it be projected down so that it reaches the ground time (51)/(6) .

A body released from the top of a tower falls through half the height of tower in 3 seconds. If will reach the ground after nearly .

The balls are released from the top of a tower of height H at regular interval of time. When first ball reaches at the ground, the n^(th) ball is to be just released and ((n+1)/(2))^(th) ball is at same distance 'h' from top of the tower. The value of h is

A body of mass m is projected horizontally with a velocity v from the top of a tower of height h and it reaches the ground at a distance x from the foot of the tower. If a second body of mass 2 m is projected horizontally from the top of a tower of height 2 h , it reaches the ground at a distance 2x from the foot of the tower. The horizontal veloctiy of the second body is.

A body of mass m is projected horizontally with a velocity v from the top of a tower of height h and it reaches the ground at a distance x from the foot of the tower. If a second body of mass 2m is projected horizontally from the top of a tower of height 2h, it reaches the ground at a distance 2x from the foot of the tower. The horizontal velocity of the second body is :