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:
$A.\,\sqrt{\dfrac{{{s}_{2}}-{{s}_{1}}}{2g}}$
$B.\,\sqrt{\dfrac{{{s}_{2}}-{{s}_{1}}}{g}}$
$C.\,\sqrt{\dfrac{{{s}_{2}}{{s}_{1}}}{g}}$
$D.\,\sqrt{\dfrac{s_{2}^{2}-s_{1}^{2}}{g}}$

Answer 1

Using the formula of the law of motion equations, this problem can be solved. This problem solution is mainly based on the formation of the equations of distances and then finding the difference between the same.

Formula used: $s=ut+\dfrac{1}{2}a{{t}^{2}}$

Complete step by step answer:
As per the given statement, a particle is freely falling under the influence of gravity, so, we will make use of the law of motion formulae to solve this problem.
The equation of the second law of motion is given as follows:
$s=ut+\dfrac{1}{2}a{{t}^{2}}$
Where s is the distance, u is initial velocity, t is the time and a is the acceleration.

This equation of the second law of motion should be expressed in terms of height reached by the particle and the acceleration should be replaced with gravitational constant.
Thus, we get,

& s=ut+\dfrac{1}{2}a{{t}^{2}} \\\ & \Rightarrow h=ut+\dfrac{1}{2}g{{t}^{2}} \\\ \end{aligned}$$ A freely falling particle under gravity covers a distance of $${{s}_{1}}$$ in first t seconds. Then u=0. $$\begin{aligned} & {{s}_{1}}=(0)t+\dfrac{1}{2}g{{t}^{2}} \\\ & \Rightarrow {{s}_{1}}=\dfrac{g{{t}^{2}}}{2} \\\ \end{aligned}$$ Again, the same freely falling particle under gravity covers a distance of $${{s}_{1}}$$ in the first t seconds. Then u=gt. $$\begin{aligned} & {{s}_{2}}=(gt)t+\dfrac{1}{2}g{{t}^{2}} \\\ & \Rightarrow {{s}_{2}}=\dfrac{3g{{t}^{2}}}{2} \\\ \end{aligned}$$ As we have obtained the two expressions of the distance at two different time intervals, so we will find the difference. So, we get, $$\begin{aligned} & {{s}_{2}}-{{s}_{1}}=\dfrac{3g{{t}^{2}}}{2}-\dfrac{g{{t}^{2}}}{2} \\\ & \Rightarrow {{s}_{2}}-{{s}_{1}}=\dfrac{2g{{t}^{2}}}{2} \\\ \end{aligned}$$ Upon the further calculation, we get, $$\begin{aligned} & {{s}_{2}}-{{s}_{1}}=g{{t}^{2}} \\\ & \Rightarrow t=\sqrt{\dfrac{{{s}_{2}}-{{s}_{1}}}{g}} \\\ \end{aligned}$$ Hence the expression for the time taken by the particle. As a particle freely falling under gravity is$$\sqrt{\dfrac{{{s}_{2}}-{{s}_{1}}}{g}}$$ **So, the correct answer is “Option A”.** **Note:** The things to be on your finger-tips for further information on solving these types of problems are: As in this problem, a freely falling body is considered, thus the acceleration should be replaced with the force of gravity, because, a freely falling body will be under the influence of the gravity. Even the displacement should be replaced with the height, as the horizontal distance is the same as that of the vertical height.