Question

A particle is falling freely under gravity. If in the first $t$ seconds, it covers distance $x_1$ and in the next $t$ seconds, it covers distance $x_2$, then $t$ is given by

• A. $\sqrt{\frac{x_2-x_1}{g}}$
• B. $\sqrt{\frac{x_2+x_1}{g}}$
• C. $\frac{\sqrt{2(x_2-x_1)}}{g}$
• D. $\frac{\sqrt{2(x_2+x_1)}}{g}$

Answer 1

The correct option is A $\sqrt{\frac{x_2-x_1}{g}}$

Taking downward as positive,
Distance covered in $1\text{st}$ $t\text{sec}$,
$x_1=\frac{1}{2}g{t}^2$
Total distance covered in $1\text{st}$ and $2\text{nd}$ $t\text{sec}$,
$x_1+x_2=\frac{1}{2}g{\left(2t\right)}^2$
$x_2=\frac{1}{2}g{\left(2t\right)}^2-\frac{1}{2}g{t}^2$
$x_2=\frac{3}{2}g{t}^2$
$x_2-x_1=g{t}^2$
$\Rightarrow t=\sqrt{\frac{x_2-x_1}{g}}$