Question

A freely falling body takes t seconds to travel first ${(\dfrac{1}{x})^{th}}$ distance, then time of descent is
A. $\dfrac{t}{{\sqrt x }}$
B. $t\sqrt x$
C. $\dfrac{{\sqrt x }}{t}$
D. $\dfrac{1}{{t\sqrt x }}$

Answer 1

In order to solve this question, we should know that, time of descent means time taken by the body to fall on ground, and free falls means body is falling under acceleration due to gravity, here we will use the general equation of motions to calculate the time taken by body to fall on ground.

Formula used:
If $u, t, g, S$ be the initial velocity, time taken, acceleration due to gravity, Distance then by equation of motion we have,
$S = ut + \dfrac{1}{2}g{t^2}$

Complete step by step answer:
According to the question, we have body is falling freely so its initial velocity be $u = 0$ and let S be the total distance to be covered by the body so one by x of total distance will be $\dfrac{S}{x}$ and time taken to cover this distance is given as t so, using
$S = ut + \dfrac{1}{2}g{t^2}$
$\Rightarrow \dfrac{S}{x} = 0 + \dfrac{1}{2}g{t^2}$
$\Rightarrow S = \dfrac{x}{2}g{t^2}$
Now, let T be the time of descent so again using same equation we have,
$S = ut + \dfrac{1}{2}g{T^2}$
put value of $\Rightarrow S = \dfrac{x}{2}g{t^2}$ we get,
$\dfrac{x}{2}g{t^2} = 0 + \dfrac{1}{2}g{T^2}$
on solving we get,
${T^2} = x{t^2}$
taking square root on both sides we get,
$\therefore T = x\sqrt t$
so, the time taken by the body to fall freely on the ground will be $T = x\sqrt t$.

Hence, the correct option is B.

Note: It should be remembered that, in case of freely falling under force of gravity, the initial velocity is always zero and when body falls in downward direction the acceleration due to gravity is positive and if body is thrown against gravity its value is taken as negative.