A lift has upward acceleration aₗ=5t m s⁻². At t=0, a ball is thrown vertically upward inside it with speed 30 m s⁻¹ relative to the lift. Taking g=10 m s⁻², the maximum separation between the ball and the point of projection in the lift is:
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Question

A lift has upward acceleration aₗ=5t m s⁻². At t=0, a ball is thrown vertically upward inside it with speed 30 m s⁻¹ relative to the lift. Taking g=10 m s⁻², the maximum separation between the ball and the point of projection in the lift is:

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G1 Physics · Accepted Answer

Relative acceleration is relative acceleration = −10 − 5t. Relative velocity: relative velocity = 30 − 10t − (5/2)t². Maximum separation occurs when relative velocity=0: 30 − 10t − 2.5t² = 0 ⇒ t²+4t−12=0 ⇒ t=2 s. Relative displacement: relative displacement = 30t − 5t² − (5/6)t³. At t=2: maximum separation = 60−20−40/6 = 40−20/3 = 100/3 m.