$A$ boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can,equal to $49\; m s^{-1}$. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of $5\; m s^{-1}$ 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?

(10 S, 10 S) Initial velocity of the ball,$u = 49\; m s^{-1}$.
Acceleration due to gravity,$a = -g = -9.8\; m s^{-2}$.
Case $I$: When the lift is stationary.
Taking the upward motion of the ball,the final velocity $v$ at the highest point is $0$.
Using the first equation of motion,$v = u + at$,the time of ascent $t_a$ is:
$t_a = \frac{v - u}{a} = \frac{0 - 49}{-9.8} = 5\; s$.
Since the time of ascent equals the time of descent,the total time taken to return to the hand is $T = t_a + t_d = 5 + 5 = 10\; s$.
Case $II$: When the lift moves up with a uniform velocity of $5\; m s^{-1}$.
Since the lift moves with a uniform velocity,its acceleration is $0$. The relative velocity of the ball with respect to the boy remains $49\; m s^{-1}$ (the same as in the stationary case). Because the frame of reference (the lift) is inertial,the time taken for the ball to return to the boy's hand remains the same,which is $10\; s$.