Question

A body is thrown with velocity $20m/s$ at an angle of ${60}^{\circ }$ with the horizontal. The time gap between the two positions of body where velocity of body makes an angle of ${30}^{\circ }$ with horizontal is

• A. $\frac{2}{\sqrt{3}}s$
• B. $\frac{1}{\sqrt{3}}s$
• C. $\sqrt{3}s$
• D. $1.5s$

Answer 1

The correct option is A $\frac{2}{\sqrt{3}}s$


Let $A$ and $B$ be the two points on the trajectory where velocity makes an angle of ${30}^{\circ }$ with the horizontal.
$u=20m/s$ (given)
Since the horizontal component of velocity always remains constant
$\Rightarrow v_1\cos {30}^{\circ }=v_2\cos {30}^{\circ }=u\cos {60}^{\circ }$
$\Rightarrow v_1=v_2=\frac{u}{\sqrt{3}}..........\text{(i)}$
Using first equation of motion for vertical component of velocity from point $A$ to $B$, we get
$-v_2\sin {30}^{\circ }=v_1\sin {30}^{\circ }-g{t}_{AB}.....\text{(ii)}$
Using $\text{(i)}$ and $\text{(ii)}$ we get
$g{t}_{AB}=2\times \frac{u}{\sqrt{3}}\times \frac{1}{2}=\frac{20}{\sqrt{3}}$
$\Rightarrow t_{AB}=\frac{2}{\sqrt{3}}s$
$\therefore$ Time gap between positions $A$ and $B$= $\frac{2}{\sqrt{3}}s$