Question

Sachin is standing on a stationary lift which is open from above . If lift starts moving up with a uniform speed of 5 m/s² and he throws the ball up with a speed equal to 147 m/s² , how much time does the ball take to return to his hands ? ( g = 9.8 m/s² )

Answer 1

Dear Student,

Initial velocity, u = 147 m/s

Now the lift starts moving up with constant velocity, v_lift = 5 m/s

Relative initial velocity of the ball w.r.t. the boy is,

u^/ = net upward velocity of the ball – net upward velocity of the boy = (147 + 5) – 5 = 147 m/s

Relative final velocity of the ball w.r.t. the boy (as the ball comes back to the hand) is,

v^/ = net upward velocity of the ball – net upward velocity of the boy

=> v^/ = (-147 + 5) – 5 = -147 m/s

The acceleration due to gravity is same, g = -9.8 m/s²

The time taken by the ball to come back to hand is, t = (v – u)/g

=> t = (-147 – 147)/(-9.8)

=> t = 30 s