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
(Take $g=10m/s^2$)

• A. $1.15s$
• B. $0.95s$
• C. $1s$
• D. $1.5s$

Answer 1

The correct option is A $1.15s$


Let $A$ and $B$ be the two points on the trajectory where velocity makes an angle of ${30}^{\circ }$ with the horizontal.
Let $v_1$ and $v_2$ be the velocities at point $A$ and $B$ respectively.
$u=20m/s$ (given)
Since the horizontal component of velocity always remains constant
$\Rightarrow v_1\cos {30}^{\circ }=v_2\cos {30}^{\circ }=u\cos {60}^{\circ }$
$\Rightarrow v_1=v_2=\frac{u}{\sqrt{3}}$ ......(i)
Using first equation of motion for vertical component of velocity from point $A$ to $B$, we get
$-v_2\sin {30}^{\circ }=v_1\sin {30}^{\circ }-g{t}_{AB}$ ......(ii)
Using (i) and (ii) we get
$g{t}_{AB}=2\times \frac{u}{\sqrt{3}}\times \frac{1}{2}=\frac{20}{\sqrt{3}}\Rightarrow t_{AB}=\frac{2}{\sqrt{3}}=1.15s$
$\therefore$ Time gap between positions $A$ and $B$ $=1.15s$