Question

From the top of a tower a stone is thrown up and reaches the ground in time $t_1$ = 9s. A second stone is thrown down with the same speed and reaches the ground in time $t_2$ = 4s. A third stone is released from rest and reaches the ground in time $t_3$, which is equal to

• A. 6.5 s
• B. 6 s
• C. $\frac{5}{36}s$
• D. 65 s

Answer 1

Answer: (B)

Taking downward motion of the first stone from A to ground, we have
$h=-ut_1+\frac{1}{2}gt^2_1$.....(i)
Taking downward motion of second stone from A to ground, we have
$h-ut_2+\frac{1}{2}gt^2_2$....(ii)
Third stone
$h=\frac{1}{2}gt^2_3$.....(iii)
Multiplying Eq. (i) by $t_2$ and Eq. (ii) by $t_1$ and adding, we get
$h(t_1+t_2)=\frac{1}{2}gt_1t_2(t_1+t_2)$
$\Rightarrow h=\frac{1}{2}gt_1t_2$.......(iv)
From Eqs. (iii) and (iv),$ t^2_3 = t_1t_2$
or $t_3\sqrt{t_1t_2}$
$=\sqrt{9 \times4 }=6s$