Question

A ball is thrown vertically upward from the roof of a building with a certain velocity. It reaches the ground in $9\, s$. When it is thrown downward from the roof with the same initial speed, it takes $4\, s$ 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

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 $\frac{1}{2} g t^{2}=\frac{1}{2} g t_{1} t_{2}$
$\therefore t=\sqrt{t_{1} \times t_{2}}$
$=\sqrt{9 \times 4}=6\, s$