Answer

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**Hint:**To solve this problem, use the equations of motions. Firstly, use the third equation of motion which describes the relation between displacement and velocity. Substitute the values in the equation and obtain the velocity of stone while passing through the window. Then, use the second equation of motion which gives the relation between time and displacement. Substitute the values in this equation and find the value of h which gives the height at which the stone dropped above the top of the window.

**Complete answer:**

Let the initial velocity of the stone be u

the velocity of the stone while passing through the window be v

the height above the window be h

the height of the window be s

time taken by the stone to pass by the window be t

Given: t= 0.2s

s= 1m

a= $10{m}/{s}$

Initial velocity of the stone was $0 {m}/{s}$.

The third equation of motion is given by,

${v}^{2} ={u}^{2} + 2ah$

Substituting u=0 we get,

${v}^{2} = 2ah$

$\Rightarrow v= \sqrt {2ah}$ …(1)

Now, the second equation of motion for the motion of stone while passing through the window is given by,

$s= vt + \dfrac {1}{2}a{t}^{2}$

Substituting values in above equation we get,

$1= \sqrt {2\times 10 \times h}\times 0.2+ \dfrac {1}{2}\times 10 \times{0.2}^{2}$

$\Rightarrow 1= \sqrt {20 \times h}\times 0.2+ \dfrac {1}{2}\times 10 \times 0.04$

$\Rightarrow 1= \sqrt {20 \times h}\times 0.2+ 0.2$

$\Rightarrow 1-0.2=\sqrt {20 \times h}\times 0.2$

$\Rightarrow \dfrac {0.8}{0.2}= \sqrt {20 \times h}$

$\Rightarrow 4= \sqrt {20 \times h}$

Squaring both the sides we get,

$20h= 16$

$\Rightarrow h=0.8m$

Hence, the stone was dropped 0.8m above the window.

**So, the correct answer is option C i.e. 0.8m.**

**Note:**

To solve these types of problems, students should know all the equations of motions. They should also know the proper application of these equations. To choose the correct equation for the specified problem, check which parameters are given and which you are asked to find. If you want to find the displacement at a given time then use the equation, $s= ut + \dfrac {1}{2}a{t}^{2}$

If you have to find the displacement at given velocity, then use the relation, ${v}^{2} ={u}^{2} + 2as$.

If you have to find the velocity at a given time then use the relation, $v = u +at$.

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