Question

A stone is allowed to fall from the top of a tower $100\text{m}$ high and at the same time another stone is projected vertically upwards from the ground with a velocity of $25\text{m/s}$. The two stones will meet at a height

• A. $21.6\text{m}$ from the ground.
• B. $50.0\text{m}$ from the ground.
• C. $78.4\text{m}$ from the ground.
• D. $19.6\text{m}$ from the ground.

Answer 1

The correct option is A $21.6\text{m}$ from the ground.

Let $u_1$ and $u_2$ be the initial velocities of the stone projected from top and bottom respectively.
$u_1=0$ and $u_2=25\text{m/s}$
$u_{21}=25\text{m/s}$
$S_{21}=100\text{m}$
Since both have acceleration in $-$ive $y$-direction
$a_{rel}=-g-(-g)=0{\text{ms}}^{-2}$
Using second equation of motion, we get
$S_{21}=u_{21}t+\frac{1}{2}{a}_{rel}{t}^2$
$100=25\text{t}$
$\Rightarrow t=4\text{s}$
Time taken by the two stones to meet $=4\text{s}$
Height from the ground at which two stones meet is given by
$h=u_{2}t-\frac{1}{2}\times g\times t^2$
$h=25\times 4-\frac{1}{2}\times 9.8\times 4^2$
$\Rightarrow h=21.6\text{m}$