Question

A ball dropped from one metre above the top of a window, crosses the window in $t_{1}s$ . If the same ball is dropped from $2m$ above the top of the same window, time taken by it to cross the window is $t_{2}s$ . Then

Answer 1

Solution 1

The ball which is dropped from greater height above the top of the window will have higher speed while crossing the window, by 3rd equation of motion:

$v^2=u^2+2aS$, where $u=0$

The ball with higher speed will take less time to cross the window.

$\because$ $H_2>H_1$

$⟹$ $v_2>v_1$

$\therefore$ $t_2<t_1$

Hence Option D is correct

Solution 2

Statement D is correct
Situation 1:
We want time to cross x.
$v^2=u^2+2as$
$u=0,a=-g,s=-1$
$v^2=2g\Rightarrow v_1=\sqrt{2g}$
$(1+x)=\frac{1}{2}g{t}^2\Rightarrow T_2=\sqrt{\frac{2(1+x)}{g}}$
$1=\frac{1}{2}g{t}^2=T_1=\sqrt{\frac{2}{g}}$
So time taken to cross $x=t_1=T_2-T_1=\sqrt{\frac{2}{g}}(\sqrt{1+x}-1)$
Situation 2:
$(2+x)=\frac{1}{2}g{t}^2\Rightarrow T_2=\sqrt{\frac{2(2+x)}{g}}$
$2=\frac{1}{2}g{t}^2\Rightarrow T_1=\sqrt{\frac{2\times 2}{g}}$
So time taken to cross $x=t_2=T_2-T_1=\sqrt{\frac{2}{g}}(\sqrt{2+x}-\sqrt{2})$
Clearly $t_2<t_1$