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 $25{\text{ms}}^{-1}$. The two stones will meet after

• A. $4\text{s}$
• B. $0.4\text{s}$
• C. $0.04\text{s}$
• D. $40\text{s}$

Answer 1

The correct option is A $4\text{s}$

Ordinary method: Using kinematics:

Let the distance from top of the tower where the stones meet be $x$,

For first stone taking downward as positive we have from the equation of motion

$S=ut+\frac{1}{2}g{t}^2$

Here, $S=x$ and $u=0$

So, for the first stone the point where the two stones meet is at

$\begin{array}{}x=\frac{1}{2}g{t}^2& \text{(1)}\end{array}$

Fo the second stone if we take the upward direction to be positive we have

$S=ut+\frac{1}{2}g{t}^2$

Here $S=100-x$ and $u=25\text{m/s}$

For the second stone the point where the two stones meet is at

$\begin{array}{}100-x=25t-\frac{1}{2}g{t}^2& \text{(2)}\end{array}$
Adding $\left(1\right)$ and $\left(2\right)$, we get

$25t=100$ or $t=4\text{s}$

Alternative method: Using Relative velocity

Initial velocity of stone $1$ which is dropped from the top of a tower, $u_1=0$

Initial velocity of stone $2$ which is thrown upward from the ground, $u_2=+25\text{m/s}$
(Take upward direction to be positive)

Hence,
$u_{rel}=u_{21}=u_2-u_1=25-0=25\text{m/s}$

$s_{rel}=100\text{m}$

$a_{rel}=g-g=0$

Using second equation of motion, we get

$S_{rel}=u_{rel}t+\frac{1}{2}{a}_{rel}{t}^2$

$\Rightarrow 100=25\times t+0$

$\therefore t=4\text{s}$

Hence, the two stones will meet after $4\text{seconds}$.