Question

A body is dropped from the top of a tower of height $3h$. The ratio of the intervals of time to cover the three equal heights $h$ is

• A. $t_1:t_2:t_3=1:3:5$
• B. $t_1:t_2:t_3=1:2:5$
• C. $t_1:t_2:t_3=\sqrt{3}:\sqrt{2}:1$
• D. $t_1:t_2:t_3=1:(\sqrt{2}-1):(\sqrt{3}-\sqrt{2})$

Answer 1

The correct option is D $t_1:t_2:t_3=1:(\sqrt{2}-1):(\sqrt{3}-\sqrt{2})$

Using first equation of motion, $v=u+at$ and using the initial condition $u=0$ we can write all three intervals in terms of respective heights as,

$t_1=\sqrt{\frac{2h}{g}}$

$t_2=\sqrt{\frac{4h}{g}}-\sqrt{\frac{2h}{g}}=\sqrt{\frac{2h}{g}}(\sqrt{2}-1)$

$t_3=\sqrt{\frac{6h}{g}}-\sqrt{\frac{4h}{g}}=\sqrt{\frac{2h}{g}}(\sqrt{3}-\sqrt{2})$

Therefore, the required ratio of time intervals is,

$t_1:t_2:t_3=1:(\sqrt{2}-1):(\sqrt{3}-\sqrt{2})$