Question

A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to $49m{s}^{-1}.t$, is the time the ball takes to return to his hands. If the lift starts moving up with a uniform speed of $5m{s}^{-1}$ and.the boy again throws the ball up with the maximum speed he can, now the ball returns to his hand in $t_2$ seconds. (Take $g=9.8m{s}^{-2}$ )

• A. $t_1=10s,t_2=10s$
• B. $t_1=10s,t_2=8s$
• C. $t_1=8s,t_2=10s$
• D. $t_1=9s,t_2=6s$

Answer 1

Answer: (A)

Using $s=ut+\frac{1}{2}a{t}^2$
$s=0,u=49m{s}^{-1},a=-9.8m{s}^{-2}$
$0=49t-\frac{9.8}{2}{t}^2$
$0=t(49-4.9t)$
$\Rightarrow t=0$ or $t=10s$
The ball comes back after $10s$. If we fix our frame of reference to the lift, the lift's initial speed will not change the answer got in the previous case. So, $t_1=t_2=10s$.