Question

A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to $49\, ms ^{-1} . t$, is the time the ball takes to return to his hands. If the lift starts moving up with a uniform speed of $5\, ms ^{-1}$ 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 $g = 9.8\, ms ^{-2}$ )

• A. $t_{1}=10\, s , t_{2}=10\, s$
• B. $t_{1}=10\, s , t_{2}=8\, s$
• C. $t_{1}=8\, s , t_{2}=10\, s$
• D. $t_{1}=9\, s , t_{2}=6\, s$

Answer 1

Answer: (A)

Using $s=u t+\frac{1}{2} a t^{2}$
$s=0, u=49\, ms ^{-1}, a = -9.8 \,ms ^{-2}$
$0 = 49 t-\frac{9.8}{2} t^{2}$
$0 = t(49-4.9 t)$
$\Rightarrow t=0$ or $t=10 \,s$
The ball comes back after $10 s$. 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, $t_{1}=t_{2}=10 s$.