Question

*
The window of the fourth floor of SANKALP building is 5 m high. A man looking out of the
window sees an object moving up and down the height of window for 2 sec. Find the height that the
object reaches from the top end of the window.​

Answer 1

Taking acceleration due to gravity, g = 10m/s^2

I have drawn the diagram for you.

Here, we should find x.

Given, t = 2s

So, Using equations of motion,

$t = \frac{2u}{g}$

$u = 10 \times 2 \div 2$

$u = 10 \frac{m}{s}$

Max. Height

$x + 5 \: = \frac{ {u}^{2} }{2g}$

$= \frac{100}{20}$

$= 5m$

That is, x = 0m.

[THANK YOU]