Question

Two particle held at different height $a$ and $b$ above the ground are allowed to fall from rest. The ratio of their velocities on reaching the ground is

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

Answer 1

The correct option is B $\sqrt{a}:\sqrt{b}$

Let we assume that both particles takes time $t_1$ & respectively to reach the ground.

$S=ut+\frac{1}{2}a{t}^2$

$g=10m/s$

For first particle, $a=5t_1^2$ ...(1)

For second particle, $b=5t_2^2$ ...(2)

By dividing $e{q}^n\left(1\right)/e{q}^n\left(2\right)$

$\frac{t_1}{t_2}=\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$

Now applying, $V=u+at$

$10t_1=V-1$ ...(3)

$10t_2=V_2$ ...(4)

Dividing $e{q}^n\left(3\right)/\left(4\right)$

$\frac{V_1}{V_2}=\frac{t_1}{t_2}=\frac{\sqrt{a}}{\sqrt{b}}$

$V_1:V_b=\sqrt{a}:\sqrt{b}$