Question

A stone is dropped from the top of building and at the same time a second stone is thrown vertically upward from the bottom of the building with a speed of $20{\text{ms}}^{-1}.$ They pass each other $3\text{seconds}$ later. Find the height of the building.
[Take $g=10{\text{m/s}}^2$]

• A. $40\text{m}$
• B. $60\text{m}$
• C. $65\text{m}$
• D. $80\text{m}$

Answer 1

The correct option is B $60\text{m}$

As per the question, both the stones will meet after $3\text{sec}$.
Let, total height of the building be $H$ and the distance covered by dropped and thrown stone be $H_1$ and $H_2$ respectively, we have
$H=H_1+H_2$
So, from equation of motion we have
$s=ut+\frac{1}{2}a{t}^2$
$\Rightarrow H_1=0\left(3\right)+\frac{1}{2}\times 10\times (3{)}^2$
$\Rightarrow H_1=45\text{m}$
also, $H_2=\left(20\right)\left(3\right)-\frac{1}{2}\times 10\times (3{)}^2$
$\Rightarrow H_2=15\text{m}$
Hence, total height of the building is
$H=45+15=60\text{m}$