A ball is projected horizontally from a table such that it collides with wall then with ground. If after collision component of velocity perpendicular to the surface is reversed in direction without change in magnitude and component of velocity parallel to the surface remains unchanged. At what distance from the wall does the ball collide with the ground. (H = 500 m, v₀ = 20 m/s, D = 100 m)

Question

G1 Physics · Accepted Answer

The ball takes t₁ = D/v₀ = 100/20 = 5 s to reach the wall. In that time it falls ½gt₁² = 125 m. At the smooth vertical wall, only the horizontal component reverses; the vertical component is unchanged. The total time to fall H = 500 m is T = √(2H/g) = 10 s, leaving 5 s after the bounce. In those 5 s the reversed horizontal velocity carries it 20 × 5 = 100 m back from the wall. Hence it hits the ground 100 m from the wall, option A.