Question

A particle is dropped from the top of a building . When it crosses a point $5\text{m}$ below the top , another particles starts to fall from a point $25\text{m}$ below the top , both particles reach the bottom of the building simultaneously. The height of the building is :
$[g=10{\text{m/s}}^2]$

• A. $45\text{m}$
• B. $35\text{m}$
• C. $25\text{m}$
• D. $50\text{m}$

Answer 1

The correct option is A $45\text{m}$


Let the speed of the particle 1 be $u_1$ , $5$ $\text{m}$ below the top of the building .
Using $v^2$ $-$ $u^2$ $=$ $2as$

$v^2$ $=$ ( $2$ ) ( $10$ ) ( $5$ ) [ $u$ $=$ $0$ ]

$v$ $=$ $10$ $\text{m/s}$

For particle $1$ using second kinematic equation we have :

$20$ + $h$ $=$ $10$$t$ + $g$ $t^2...........\left(1\right)$

For particle 2 , using second kinematic equation we have :

$h$ $=$ $g$ $t^2.............\left(2\right)$

Using equations $\left(1\right)$ and $\left(2\right)$ we have

$20$ + $g{t}^2$ $=$ $10t$ + $g{t}^2$

$t$ $=$ $2$ $\text{sec}$

Using this value in equation $\left(1\right)$ we have
$h$ $=$ $20$ $\text{m}$

The height of the building is,
$25+20$ $=$ $45\text{m}$

Hence, $\left(A\right)$ is the correct answer.