Question

A stone falls freely from rest from a height $h$ and it travels a distance $\dfrac{{9h}}{{25}}$ in the last second. The value of $h$ is:
A. $145m$
B. $100m$
C. $122.5m$
D. $200m$

Answer 1

The Earth attracts every object towards its center applying the force of gravity. Due to this acceleration, a constant acceleration is produced called acceleration due to gravity, denoted by $g$. So, during freefall only gravity acts on the object, all other forces are assumed to be negligible.

Complete step by step solution:
Let us first write the information given in the question.
Stonefalls from rest, so initial velocity will be zero $u = 0$, distance traveled in the last second $s = \dfrac{{2h}}{{25}}$ and we have to calculate the height $h$.
We have the following formula to calculate the distance.
$s = ut + \dfrac{1}{2}a{t^2}$
Here, $s$is the distance traveled, $u$ is the initial velocity, $a$ is the acceleration, and $t$ is the total time taken.
Now, let $h$be the total height traveled in time $t$, so the above equation reduces to the following form.
$h = \left( 0 \right)t + \dfrac{1}{2}\left( g \right){t^2} \Rightarrow h = \dfrac{{g{t^2}}}{2}$ ……………………..(1)
Similarly, the distance traveled in time $\left( {t - 1} \right)$ second is given below.
$h' = \left( 0 \right)t + \dfrac{1}{2}\left( g \right){\left( {t - 1} \right)^2} \Rightarrow h' = \dfrac{{g{{\left( {t - 1} \right)}^2}}}{2}$…………………….(2)
Now the distance traveled in the last second is calculated by subtracting the distance traveled in $t$second and distance traveled in $\left( {t - 1} \right)$ second.
$s = \dfrac{{g{t^2}}}{2} - \dfrac{{g{{\left( {t - 1} \right)}^2}}}{2} = \dfrac{{g{t^2} - g{t^2} - g + 2gt}}{2} = \dfrac{{g\left( {2t - 1} \right)}}{2}$
Let us substitute the value in the above equation.
$\dfrac{{2h}}{{25}} = \dfrac{{g\left( {2t - 1} \right)}}{2}$
$\Rightarrow h = \dfrac{{25g}}{4}\left( {2t - 1} \right)$……………….(3)
Now, equating equations (1) and (3).
$\dfrac{{25g}}{4}\left( {2t - 1} \right) = \dfrac{{g{t^2}}}{2}$
Let us simplify the above expression.
$2{t^2} - 50t + 25 = 0$
On solving this quadratic equation, we get the following.
$t = 5\sec$
And, $t = \dfrac{5}{9}\sec$
Now, we will take $t = 5\sec$ as it is mentioned last second.
Now let us put this value of time in equation (1).
$h = \dfrac{{\left( {9.8} \right){{\left( 5 \right)}^2}}}{2} = \dfrac{{245}}{2} = 122.5m$
Hence, the correct option is (C) $122.5m$.

Note:
Whenever questions say free fall, it means only acceleration due to gravity is acting on the body.
The value of acceleration due to gravity is $9.8m/{s^2}$.
During freefall, the initial velocity will always be zero.