A projectile is launched from a horizontal floor with speed 20 m/s at 45°. It collides with the same smooth floor with coefficient of restitution e = 1/2. Take g = 10 m/s². If the horizontal distance between the first and second impacts is N m, find N.
Physics · Center of maas , Momentum and Collision · collision and types of collision
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Question

A projectile is launched from a horizontal floor with speed 20 m/s at 45°. It collides with the same smooth floor with coefficient of restitution e = 1/2. Take g = 10 m/s². If the horizontal distance between the first and second impacts is N m, find N.

Physics · Center of maas , Momentum and Collision · collision and types of collision

G1 Physics · Accepted Answer

Initial components: u_x = 20cos45° = 10√2 m/s, u_y = 20sin45° = 10√2 m/s. The first flight time is T1 = 2u_y/g = 2(10√2)/10 = 2√2 s. Hence first range is R1 = u_xT1 = (10√2)(2√2) = 40 m. Just before first impact, vertical speed magnitude is 10√2 m/s. After impact, upward vertical speed becomes e(10√2) = 5√2 m/s. Second-flight time: T2 = 2(5√2)/10 = √2 s. Horizontal speed remains 10√2 m/s. Therefore distance between first and second impacts: R2 = (10√2)(√2) = 20 m. Hence N = 20.