Question

A ball is thrown vertically upward from the roof of a building with a certain velocity. It reaches the ground in $9s$. When it is thrown downward from the roof with the same initial speed, it takes $4s$ to come at ground. How much time (in second) will it take to reach at ground if it just released from the rest from the roof?

Answer 1

Answer: 6

Let the height of building is $h$.
Event $I,-h=v_0{t}_1-\frac{1}{2}g{t}_1^2$ ...(i)
Event II, $h=v_0{t}_2+\frac{1}{2}g{t}_2^2$ ...(ii)
Event III, $h=\frac{1}{2}g{t}^2$ ...(iii)
On multiplying Eq. (i) by $t_2$ and Eq. (ii) by $t_1$ and add, we get
$h\left(t_1+t_2\right)=\frac{1}{2}g{t}_1{t}_2\left(t_1+t_2\right)$
Or $\frac{1}{2}g{t}^2=\frac{1}{2}g{t}_1{t}_2$
$\therefore t=\sqrt{t_1\times t_2}$
$=\sqrt{9\times 4}=6s$