A falling stone takes $0.2$ seconds to fall past a window which is $1$m high. From how far above the top of the window was the stone dropped?
A. $0.8$m
B. $0.6$m
C. $0.4$m
D. None of these.

Hint: Basic concept of equation of motion is to be used. As the stone is falling under gravity alone, so acceleration $ = g = 10m/{s^2}$
1. $s = ut + \dfrac{1}{2}a{t^2}$
2. \[{v^2} - {u^2} = 2as\]

Complete step by step answer:
When the stone was dropped, its initial velocity was zero.
Let at the start of window, the stone has velocity, v
Now, the window is $1m$ high. The distance covered by stone to part the window, $s = 1m$ and time taken for it, $t = 0.2\sec $. By equation of instion, we know that \[s = ut + \dfrac{1}{2}a{t^2}\]
Where S is distance
U is initial velocity $ = {v_1}$
T is the take
a - acceleration $ = g = 10m/{s^2}$
Putting the values, we get
\[I = vt + \dfrac{1}{2}g{t^2}\]
$I = v\left( {0.2} \right) + \dfrac{1}{2} \times 10 \times {\left( {0.2} \right)^2}$
$I = 0.2v + \dfrac{{10 \times 0.04}}{2}$
$ \Rightarrow 1 = 0.2v + 0.2$
$ \Rightarrow 1 - 0.2 = 0.2v$
$ \Rightarrow 0.8 = 0.2v$
$ \Rightarrow v = \dfrac{{0.8}}{{0.2}}$
$ \Rightarrow v = 4m/s$
Initially, from the point where it is dropped, the initial velocity, $u = 0\,\,m/s$
So, using equation
${v^2} - {u^2} = 2as$
${4^2} - {0^2} = 2 \times 10 \times s$
$16 = 20 \times s$
$s = \dfrac{{10}}{{20}}$
$s = \dfrac{8}{{10}}$
$ \Rightarrow s = 0.8m$
So, the ball is dropped $0.8m$ above the top of the window.

So, the correct answer is “Option A”.

Note:
As we need to find the distance above top of the window from where the stone is dropped. So, that is why the velocity appearing at top of window is taken as the final velocity $ = v = 4m/s$ and for free fall, initial velocity, $u = 0\,\,m/s$. Also for free fall, acceleration $ = $ acceleration due to gravity.