Question

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

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

Answer 1

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


Horizontal component of velocity of particle remains constant during projectile motion.
Hence we can equate the horizontal component of velocity of body at point $O$ and $A$
$u\cos {60}^o=v\cos {30}^o$
$v=\frac{20}{\sqrt{3}}m/s$
We can assume the motion of body between points $A$ and $B$ as projectile in horizontal plane
Hence, from the formula of time of flight
$T=\frac{2v\sin {30}^o}{g}=2\times \frac{20}{\sqrt{3}}\times \frac{1}{20}$
$T=\frac{2}{\sqrt{3}}s$