Question

A stone is dropped from the top of a building. When it crosses a point $5m$ below the top, another stone starts to fall from a point $25m$ below the top, both stones reach the bottom of building simultaneously. The height of the building is: $\left[Takeg=10\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.\right]$

• A. $45m$
• B. $35m$
• C. $25m$
• D. $50m$

Answer 1

The correct option is A

$45m$

Step 1: Given data assumptions.

Height covered by the first stone$=5m$

Height covered by the second stone$=25m$

Acceleration due to gravity, $g=10\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$

Step 2: Finding The height of the building.

Formula Used:

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

Where, $S$ is distance, $u$ initial velocity, and $t$ is time

Since,

The velocity of the first stone at $5m$ below top, $u_1=\sqrt{2gh}=\sqrt{2\times 10\times 5}=10\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Now,

Apply a formula for the first stone which is shown in the figure we get,

$20+h=u_{1}t+\frac{1}{2}g{t}^2$

$\Rightarrow$$20+h=10t+\frac{1}{2}g{t}^2$ …..$\left(i\right)$

Apply a formula for the second stone which is shown in the figure we get,

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

$\Rightarrow$$h=\frac{1}{2}g{t}^2$ $\left[\because u=0\right]$ …..$\left(ii\right)$

Substitute equation $\left(ii\right)$ in $\left(i\right)$ we get.

$20+\frac{1}{2}g{t}^2=10t+\frac{1}{2}g{t}^2$

$\Rightarrow$ $t=2s$ ….$\left(iii\right)$

Now, again substitute equation $\left(iii\right)$ in equation $\left(ii\right)$ we get.

$h=\frac{1}{2}\times 10\times 2^2$

$\Rightarrow$$h=20m$

Hence, the height of the building from the drawn figure$=h+25$$=20+25=45m$

Therefore, option A is correct.