Question

From the foot of a tower $90\text{m}$ high, a stone is thrown up so as to reach the top of the tower with zero velocity. Two seconds later another stone is dropped from the top of the tower. Find when will two stones meet for the first time ?

• A. $4.5\text{s}$
• B. $1.7\text{s}$
• C. $3.1\text{s}$
• D. $2.8\text{s}$

Answer 1

The correct option is C $3.1\text{s}$

First stone is thrown so as to reach the top of the tower, so its initial velocity is given by
$v^2-u^2=2(-g)H$ (By third equation of motion)
$\Rightarrow 0-u^2=-2gH$
$\Rightarrow u=\sqrt{2gH}$
$=\sqrt{2\times 10\times 90}=42.5\text{m/s}$
Let us take the time $t=t_0$ when the two stones meet at a height $H$ above the foot of the tower. The first stone is travelled a height $H$ in the duration $t_0$ and the second stone has fallen a distance $(90-H)$ in time $(t_0-2)$.
Using second equation of motion
For first stone : $H=42.5t_0-\frac{1}{2}\left(10\right)t_0^2$
For second stone: $90-H=\frac{1}{2}\left(10\right)(t_0-2{)}^2$
Adding above two equation, we get
$22.5t_0=70$
Or $t_0=3.11\text{s}$