Question

A body dropped from a height “$h$” at time $t = 0.$ ,reaches the ground at time ${t_o}$ it would have reached a height $\dfrac{h}{2}$ at time $t = ........$
A. $\dfrac{{{t_o}}}{{\sqrt 2 }}$
B. ${t_0}$
C. $\dfrac{{{t_o}}}{3}$
D. None of these

Answer 1

In order to solve this question, we should know that when a body is dropped, it falls freely under the force of gravity and its initial velocity is zero, we will use the newton’s equation of motion to derive a relation between time taken and distance covered by the body and then we will find time taken by the body to cover a distance of $\dfrac{h}{2}$

Formula Used:
When a body falls freely under force of gravity, its acceleration is the acceleration due to gravity which is $g = 9.8\,m{s^{ - 2}}.$
Newton’s equation of motion $S = ut + \dfrac{1}{2}a{t^2}$.
Where $u,t,a{\kern 1pt} {\kern 1pt} {\kern 1pt} and{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} S$ are the initial velocity, time taken, acceleration and Distance covered by the body during the journey.

Complete step by step answer:
According to the question, we have given that body is dropped at $t = 0.$ freely so,
Initial velocity of the body, $u = 0$.
Total distance covered by the body which is height h, $S = h$.
Total time to cover height h, $t = {t_o}$.
Acceleration of the body, $a = g$.
Now, putting these values in formula we have,
$S = ut + \dfrac{1}{2}a{t^2}$
$\Rightarrow h = 0 + \dfrac{1}{2}g{t_o}^2$
$\Rightarrow 2h = g{t_o}^2 \to (i)$
Now, in order to cover a distance of height $\dfrac{h}{2}$ let us assume body takes time $t'$ then again, using the relation,
$S = ut + \dfrac{1}{2}a{t^2}$
We get,
$\Rightarrow \dfrac{h}{2} = 0 + \dfrac{1}{2}gt{'^2}$
$\Rightarrow h = gt{'^2}$
From equation (i) put the value of $h = \dfrac{{g{t_o}^2}}{2}$ in above equation we get,
$\dfrac{{g{t_o}^2}}{2} = gt{'^2}$
$\therefore t' = \dfrac{{{t_o}}}{{\sqrt 2 }}$
So, body will take time $t' = \dfrac{{{t_o}}}{{\sqrt 2 }}$ to cover a height of $\dfrac{h}{2}$.

Hence, the correct option is A.

Note: It should be remembered that, while solving such questions if body is dropped freely its initial velocity is always zero and acceleration is acceleration due to gravity which is positive taken but if body is thrown upwards against gravity then acceleration due to gravity will be taken as negative value. The magnitude of Acceleration due to gravity is always different for different planets and places, for earth its value is taken as $g = 9.8\,m{s^{ - 2}}.$