Problem

Problem

An open elevator is ascending with constant speed v=10 m/sv = 10 \text{ m/s} . A ball is thrown vertically up by a boy on the lift when he is at a height h=10 mh = 10 \text{ m} from the ground. The velocity of projection is v=30 m/sv = 30 \text{ m/s} with respect to elevator. Find (a) the maximum height attained by the ball. (b) the time taken by the ball to meet the elevator again. (c) time taken by the ball to reach the ground after crossing the elevator.
Intuitive explanation

The key insight is that the ball's motion has two independent parts: it inherits the elevator's constant upward velocity, then experiences projectile motion relative to the ground. Since the elevator moves at constant speed, it doesn't accelerate, so you can treat the ball's initial velocity as vball=30+10=40 m/sv_{\text{ball}} = 30 + 10 = 40 \text{ m/s} upward from height h=10 mh = 10 \text{ m}, with only gravity acting on it afterward. This transforms a tricky reference-frame problem into standard kinematics.

Intuitive explanation

The key insight is that the ball's motion has two independent parts: it inherits the elevator's constant upward velocity, then experiences projectile motion relative to the ground. Since the elevator moves at constant speed, it doesn't accelerate, so you can treat the ball's initial velocity as vball=30+10=40 m/sv_{\text{ball}} = 30 + 10 = 40 \text{ m/s} upward from height h=10 mh = 10 \text{ m}, with only gravity acting on it afterward. This transforms a tricky reference-frame problem into standard kinematics.

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