Question

$A$ , $B$ , $C$ are points on a vertical line such that $AB=BC$ . If a body is dropped from rest at $A$ , and $t_1$ and $t_2$ are the time to travel for distance $AB$ and $BC$ , then ratio $\left(\frac{t_2}{t_1}\right)$ is

• A. $\sqrt{2}+1$
• B. $\sqrt{2}-1$
• C. $2\sqrt{2}$
• D. $\frac{1}{\sqrt{2}+1}$

Answer 1

Answer: (B)

$-h=0-\frac{1}{2}g{t}^2$
$t=\sqrt{\frac{2h}{g}}\dots \dots ..\left(1\right)$
$-\frac{h}{2}=0-\frac{1}{2}g{t}_1^2$
$t_1=\sqrt{\frac{h}{g}}\dots \dots ..\left(2\right)$
$so{t}_2=t-t_1$
$=\sqrt{\frac{2h}{g}}-\sqrt{\frac{h}{g}}$
$t_2=\sqrt{\frac{h}{g}}\left(\sqrt{2}-1\right)\dots \dots .\left(3\right)$
equation $3/1$
$\frac{t_2}{t_1}=\frac{\sqrt{\frac{h}{g}}\left(\sqrt{2}-1\right)}{\sqrt{\frac{h}{g}}}$
$\frac{t_2}{t_1}=\sqrt{2}-1$