A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m/s. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of 5 m/s and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands?
Given,
Initial speed of the ball, u = 49 m/s
Acceleration, a = -g = -9.81 m/s2
Speed of the lift = 5 m/s
Case 1, Lift is stationary:
It resembles freely falling body. Thus, In upward motion final velocity, v = 0
From 1st equation of motion,
v = u + at
Where,
v = Final velocity
u = Initial velocity
a = Acceleration/Deceleration
t = Time
⇒ Time of ascent, t = 
s = 
s = 5 s
Total time of float = 2×Time of ascent = 10 s
Case 2, Lift is moving with velocity 5 m/s:
Even though, lift (boy) is moving with velocity 5 m/s the initial velocity also increased by 5 m/s. So, relative velocity of ball with respect to boy remains 49 m/s. Thus, Time of float = 10 s.