Question

A particle is falling freely under gravity from rest. In first $t$ second it covers distance $x_1$ and in the next $t$ second it covers distance $x_2,$ then $t$ is given by:

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

Answer 1

The correct option is B $\sqrt{\frac{x_2+x_1}{2g}}$

If we consider first $2t$ seconds, distance travelled is $x_1+x_2$
Hence, $v^2-u^2=2as$
Since $u=0$, $a=g$ and $s=x_1+x_2$,
$v^2=2g(x_1+x_2)$
$v=\sqrt{2g(x_1+x_2)}$
Now, $v=u+at$ gives us
$t=\frac{v}{a}=\frac{\sqrt{2g(x_1+x_2)}}{g}$
Thus, $t=\sqrt{\frac{(x_1+x_2)}{2g}}$