Question

In the last second of the free fall, a body covered $\dfrac{3}{4}th$ of its total path. Then the height from which the body released will be
A. $4.9m$
B. $9.8m$
C. $19.6m$
D. $39.2m$

Answer 1

Hint:In the question, we are provided a problem of kinematics of a freely falling body. In the question, we have to take the initial velocity to be zero. Then applying the equations of motion, we can find our solution. We have to take the last second of the fall to be $t - 1$ where t is the total time of fall.

Complete step by step answer:
Consider $h$ is the total height. For a freely falling body, we take initial velocity to be zero. So, $u = 0$.Also consider $t$ is the time of fall and the acceleration to be $g = 10m{s^{ - 2}}$. Now, applying the equations of motion:
$s = ut + \dfrac{1}{2}a{t^2}$
$
h = 0 \times t + \dfrac{1}{2}\left( {10} \right){t^2} \\
\Rightarrow h = 5{t^2} \\ $ $ \ldots \left( 1 \right)$
Now, in the last second the initial velocity $u$ becomes $u + g\left( {t - 1} \right)$ and time $t$ becomes $t - 1$ and we are provided with that the total height in the last second is $\dfrac{3}{4}th$ of total height. So, height $h$changes to $\dfrac{3}{4}h$
Now, the equation of motion be:
$\dfrac{3}{4}h = \left[ {u + g\left( {t - 1} \right)} \right]\left( {t - 1} \right) + \dfrac{1}{2}g{\left( {t - 1} \right)^2}$
Using $\left( 1 \right)$ and solving
$\dfrac{3}{4}\left( {5{t^2}} \right) = \dfrac{3}{2}\left( {10} \right){\left( {t - 1} \right)^2}$
$
\Rightarrow {t^2} = 4{\left( {t - 1} \right)^2} \\
\Rightarrow 3{t^2} - 8t + 4 = 0 \\
\Rightarrow 3{t^2} - 6t - 2t + 4 = 0 \\
\Rightarrow 3t(t - 2) - 2(t - 2) = 0 \\
\Rightarrow (t - 2)(3t - 2) = 0 \\ $
Hence, we got $t = 2,\dfrac{2}{3}$
But time cannot be in fraction. Hence, time of fall is $2\sec $
Now putting the value in $\left( 1 \right)$
$
h = 5{\left( 2 \right)^2} \\
\Rightarrow h = 20 \\ $
Total height will be $20m$ when we take the $g = 10m{s^{ - 2}}$.
And the height by taking $g = 9.8m{s^{ - 2}}$.
$\therefore h = \dfrac{1}{2} \times 9.8 \times {2^2} = 19.6$

Hence, the total height is $19.6m$. So, the correct option is C.

Note:There is big confusion when we should take the acceleration to be $9.8m{s^{ - 2}}$ or $10m{s^{ - 2}}$. Then the answer is unless we are given with the value in the question and if we are not given then It is our choice to take either value. It hardly makes a difference in our solution. Keep in mind to take the initial velocity to be zero for a freely falling body.