A man standing on the top of a tower throws a ball vertically up with a certain velocity. He also throws another ball vertically down with the same speed. Which ball will hit the ground with higher speed? Neglect air resistance.
Answer
Suppose the man standing on the top of the tower is at a height ‘h’ from the ground. Let his position is marked as ‘A’. Case 1 Suppose he throws a ball of mass ‘m’ up with a certain initial velocity v. The ball moves up with an acceleration = -g. We suppose there are no frictional forces present. Let at some height h₁, the ball comes to rest. Potential energy possessed by the ball with respect to the position of its thrower is mgh₁. The ball now starts its downward journey and this P-E is converted to Kinetic Energy, which is given by ½(mv²). As mass of the ball is constant and there is no frictional forces so reaching the point ‘A’ the ball will have the same kinetic energy. This means the velocity of the downward motion at point ‘A’ will be the same v. Case 2 The second ball is thrown downward from the same height with the same velocity v. Now the situation is like both balls are thrown downward with the same velocity, v from the same point ‘A’ at height ‘h’ from the ground. Therefore, reaching the ground both balls will have the same velocity. Thus both balls will hit the ground with the same speed.