Question

A particle is falling freely under the action of gravity. It covers a distance $x_1$ in the first ${}^{\prime }{t}^{\prime }$ seconds and distance $x_2$ in the next ${}^{\prime }{t}^{\prime }$ seconds, then ${}^{\prime }{t}^{\prime }$ is given by

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

Answer 1

The correct option is C $\sqrt{\frac{x_2-x_1}{g}}\text{sec}$

As the particle is under free fall, its initial velocity $u=0$
$\therefore$ from the equation of motion we have
$x_1=0+\frac{1}{2}g{t}^2$
and, $x_1+x_2=0+\frac{1}{2}g(2t{)}^2$
$\Rightarrow x_2=2g{t}^2-x_1=2g{t}^2-\frac{1}{2}g{t}^2=\frac{3}{2}g{t}^2$
Now, we get
$x_2-x_1=(\frac{3g{t}^2}{2}-\frac{g{t}^2}{2})=g{t}^2$
$\Rightarrow t=\sqrt{\frac{x_2-x_1}{g}}$