Question

Two bodies of different masses $m_a$ and $m_b$ are dropped from two different heights $a$ and $b.$ The ratio of the time taken by the two to cover these distances are

• A. $a:b$
• B. $b:a$
• C. $\sqrt{a}:\sqrt{b}$
• D. $a^2:b^2$

Answer 1

The correct option is C

$\sqrt{a}:\sqrt{b}$

We know initial velocity, $u$ of both the bodies is zero.
Using equation of motion, $S=ut+\frac{1}{2}a{t}^2$
where, $S=$ height of the bodies$\left(h\right),u=0$, $a=$ acceleration due to gravity$\left(g\right)$ and $t=$ time period.
we have,

$h=\frac{1}{2}g{t}^2$

$\Rightarrow t=\sqrt{\frac{2h}{g}}$
Let time for body 'a' be $t_a$ and time for body 'b' be $t_b$.
Given, height of 'a'= a and height of 'b'= b.

For the two bodies in question, we have

$t_a=\sqrt{\frac{2a}{g}}$ and $t_b=\sqrt{\frac{2b}{g}}$

$\Rightarrow \frac{t_a}{t_b}=\sqrt{\frac{a}{b}}$