A particle is projected at an angle of ${60}^{\circ }$ above the horizontal with a speed of $10m/s$. After some time, the direction of its velocity makes an angle of ${30}^{\circ }$ above the horizontal. The speed of the particle at this instant is

The correct option is D $\frac{10}{\sqrt{3}}m/s$
Horizontal component of velocity of the particle initially (at ${\theta }_1={60}^{\circ }$ ),
$v_H=u\cos {\theta }_1{v}_H=10\times \cos {60}^{\circ }=10\times \frac{1}{2}{v}_H=5m/s$
As the horizontal component of velocity remains constant throughout the projectile motion.
Let $v$ be the velocity when the particle makes an angle of ${30}^{\circ }$ with the horizontal.
So, $v_H=v\cos {\theta }_2$
$v=\frac{v_H}{\cos {30}^{\circ }}$
$=\frac{5}{\frac{\sqrt{3}}{2}}$
$\therefore v=\frac{10}{\sqrt{3}}m/s$