Question

A particle is falling freely under gravity. 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. $\sqrt{\frac{x_2+x_1}{g}}$
• C. $\sqrt{\frac{(x_2-x_1)}{2g}}$
• D. $\sqrt{\frac{(x_2+x_1)}{g}}$

Answer 1

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

$S=ut+\frac{1}{2}g{t}^2$

$x_1=ut+\frac{1}{2}g{t}^2$

$v=u+gt$

$x_2=vt+\frac{1}{2}g{t}^2=(u+gt)t+\frac{1}{2}g{t}^2=ut+\frac{3}{2}g{t}^2$

subtract $2nd$ eq from $1st$

$x_2-x_1=g{t}^2$

$t=\sqrt{\frac{x_2-x_1}{g}}$