A ball is projected horizontally with speed u from a height h above a smooth horizontal floor. The coefficient of restitution with the floor is e. The horizontal distance travelled between the first and second collisions with the floor is:
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Question

A ball is projected horizontally with speed u from a height h above a smooth horizontal floor. The coefficient of restitution with the floor is e. The horizontal distance travelled between the first and second collisions with the floor is:

Options

G1 Physics · Accepted Answer

Just before the first impact, the vertical speed is obtained from free fall: v_y = √(2gh). After the collision, the upward vertical speed becomes e√(2gh). The time from the first collision to the second collision is the complete up-and-down flight time: T = 2[e√(2gh)]/g = 2e√(2h/g). The floor is smooth, so horizontal speed remains u. Therefore horizontal distance between first and second impacts is x = uT = 2eu√(2h/g). Correct option: B.