Question

An object falls from rest from the top of a building. An observer inside the building notices that it takes $0.2s$ for the object to pass his window, whose height is $1.6m$. How far above the top of this window is the place from where object falls? (Take $g=9.8m{s}^{-2}$)

Answer 1

Freely Falling Body:

1. When a body is dropped from a height without any initial velocity given to the body under the influence of gravity of the earth said to be falling freely.
2. The initial velocity of the body is zero.
3. The acceleration applied to the body is equal to the gravity of the earth and its direction is downwards.

Step 1: Given data

Time taken, $t=0.2s$

Height of the window, $s=1.6m$

Acceleration due to gravity, $g=9.8m{s}^{-2}$

(Note: We have considered the downward direction as a positive direction).

Step 2: Calculate the initial velocity of the object

As we know, the second equation of motion under gravity is $s=ut+\frac{1}{2}g{t}^2$

$$\begin{gathered} \Rightarrow s=ut+\frac{g{t}^2}{2} \\ \Rightarrow 1.6=u(0.2)+\frac{9.8\times 0.2^2}{2} \\ \Rightarrow u=7.02m{s}^{-1} \end{gathered}$$

Step 3: Calculate the height of the window from the ground

The initial velocity of the object during the fall will be equal to the final velocity of the object when it is passing the window.

As we know, the third equation of motion under gravity is $v^2=u^2+2gs$

$$\begin{gathered} \Rightarrow v^2=u^2+2gh \\ \Rightarrow 7.{02}^2=0^2+2(9.8)\left(h\right) \\ \Rightarrow h=2.52m \end{gathered}$$

Final Answer:

Thus, the window is $\mathbf{2}\mathbf{.}\mathbf{52}\mathbf{}\mathit{m}$ above the place from where the object falls.