Question

Multiple Correct Answers Type:
Two bodies of masses $m_1$ and $m_2$ are dropped from heights $h_1$ and $h_2$, respectively. They reach the ground after time $t_1$ and $t_2$ and strike the ground with $v_1$ and $v_2$, respectively. Choose the correct relations from the following.

• A. $\frac{t_1}{t_2}=\sqrt{\frac{h_1}{h_2}}$
• B. $\frac{t_1}{t_2}=\sqrt{\frac{h_2}{h_1}}$
• C. $\frac{v_1}{v_2}=\sqrt{\frac{h_1}{h_2}}$
• D. $\frac{v_1}{v_2}=\frac{h_2}{h_1}$

Answer 1

Answer: (A), (C)

The correct options are
A $\frac{t_1}{t_2}=\sqrt{\frac{h_1}{h_2}}$
C $\frac{v_1}{v_2}=\sqrt{\frac{h_1}{h_2}}$

In free fall, initial velocity $u=0$

apply second kinematic equation

$v^2-u^2=2gh$

$v=\sqrt{2gh}$

$v\alpha \sqrt{h}$

Hence, $\frac{v_1}{v_2}=\sqrt{\frac{h_1}{h_2}}$

Apply third kinematic equation

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

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

$t\alpha \sqrt{h}$

Hence, $\frac{t_1}{t_2}=\sqrt{\frac{h_1}{h_2}}$

Relation as follows,

$\frac{v_1}{v_2}=\sqrt{\frac{h_1}{h_2}}\&\frac{t_1}{t_2}=\sqrt{\frac{h_1}{h_2}}$