Taking vertical upward direction as the positive direction of x-axis.When lift is stationary, consider the motion of the ball going verticallyupwards and coming down to the hands of the boy, we have $u=49\ \mathrm{m}\ \mathrm{s}^{-1},\ a=-9.8\ \mathrm{m}\ \mathrm{s}^{-2},$ $t=?,\ x-x_{0}=S=0$ As, $S=u t+\frac{1}{2}a t^{2}$ $\therefore\; 0=49 t+\frac{1}{2}(-9.8) t^{2}$ or $49 t=4.9 t^{2}$ or $t=\frac{49}{4.9}=10$ seconds As the lift starts moving upwards with uniform speed of 5 m s$^{-1}$, there isno change in the relative velocity of the ball with respect to the boy i.e., itremains 49 m s$^{-1}$. Hence, even in this case, the ball will return to the boy’shand after 10 second.