A ball is thrown from the top of a tower in vertically upward direction. Velocity at a point h m below the point of projection is twice of the velocity at a point h m above the point of projection. Find the maximum height reached by the ball above the top of tower.

A ball is thrown vertically up with a certain velocity from the top of a tower of height 40m. At 4.5m above the top of the tower its speed is exactly half of that it will have at 4.5m below the top of the tower. The maximum height reached by the ball aboe the ground is 47.5K m. Find K.

A particle is projected vertically upwards from an elevated point. Magnitude of the velocity at height h above the starting point is found to be half of the magnitude of velocity at h height below the starting point. If maximum height reached by the particle above its initial point is mh/n then find (m-n).

A ball thrown by one player reaches the other in 2 s . The maximum height attained by the ball above the point of projection will be about.

A ball is thrown vertically upward with a velocity u from the top of a tower. If it strikes the ground with velocity 3u, the time taken by the ball to reach the ground is

A ball is thrown horizontally from the top of a tower 40 m high. The ball strikes the ground at a point 80 m from the bottom of the tower. Find the angle that the velocity vector makes with the horizontal just before the ball hits the gound.

A ball is thrown vertically upwards from the top of tower of height h with velocity v . The ball strikes the ground after time.

A ball is thrown from the top of a tower of 61 m high with a velocity 24.4 ms^(-1) at an elevation of 30^(@) above the horizontal. What is the distance from the foot of the tower to the point where the ball hits the ground?

A ball is thrown from a point in level with velocity u and at a horizontal distance r from the top os a tower of height h . How must the speed and angle of the projection of the ball be related to r in order that the ball may just go grazing the top edge of the tower ?