Question

A stone is allowed to fall from the top of a tower $80\text{m}$ high and at the same time another stone is projected vertically upwards from the ground with a velocity of $20\text{m/s}$ along the same line of motion. The two stones will meet after -

• A. $2\text{s}$
• B. $4\text{s}$
• C. $8\text{s}$
• D. $6\text{s}$

Answer 1

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

Let the two stones meet at point $P$ as shown in the figure.


For stone allowed to fall from the top of the tower :

$u_1=0$
$a_1=-10{\text{m/s}}^2$
$s_1=-x$

Applying,

$s_1=u_{1}t+\frac{1}{2}{a}_1{t}^2$

$\Rightarrow -x=0+\frac{1}{2}(-10)t^2$

$\Rightarrow x=5t^2...\left(1\right)$

For stone projected from the bottom of the tower:

$u_2=20\text{m/s}$
$a_2=-10{\text{m/s}}^2$
$s_2=80-x$

Applying,

$s_2=u_{2}t+\frac{1}{2}{a}_2{t}^2$

$\Rightarrow 80-x=20t+\frac{1}{2}(-10)t^2$

$\Rightarrow 80-x=20t-5t^2...\left(2\right)$

Adding equations $\left(1\right)$ and $\left(2\right)$, we get,

$80=20t$

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

Therefore, the two stones will meet after $4\text{s}$.

​​​​​​​Hence, option $\left(B\right)$ is the correct answer.