Question

A particle is falling freely under gravity. In first $t$ second it covers $s_1$ and in the next $t$ seconds it covers $s_2$ then $t$ is given by :

• A. $\sqrt{\frac{s_2-s_1}{2g}}$
• B. $\sqrt{\frac{s_2-s_1}{g}}$
• C. $\sqrt{\frac{s_2{s}_1}{g}}$
• D. $\sqrt{\frac{s_2^2-s_1^2}{g}}$

Answer 1

The correct option is B $\sqrt{\frac{s_2-s_1}{g}}$

Solution In first t set,

$h=4t+\frac{a}{2}{t}^2$ [second $e{q}^n$ of motion]

$h=o\left(t\right)+\frac{a}{2}{t}^2$

$\Rightarrow s_1=\frac{g{t}^2}{2}$

In second t sec,

$g=u^{1}t+\frac{a{t}^2}{2}$

$u^1=u+at$ [first $e{q}^n$ of motion]

$u^1=\left(o\right)+gt$

So, $h=\left(gt\right)t+\frac{g}{2}{t}^2$

$\Rightarrow s_2=\frac{3g{t}^2}{2}$

So, $s_2-s_1=\frac{3}{2}g{t}^2-\frac{g{t}^2}{2}=g{t}^2$

$\Rightarrow t=\sqrt{\frac{s_2-s_1}{g}}$