Q. A particle is projected at an angle of $60{^\circ}$ above the horizontal with a speed of $10\ ms^{- 1}$. 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
A) $\frac{5}{\sqrt{3}}ms^{- 1}$
B) $5\sqrt{3}\ ms^{- 1}$
C) $5\ ms^{- 1}$
D) $\frac{10}{\sqrt{3}}ms^{- 1}$
Explanation
(d) Let $v$ the velocity of projectile at this instants. Horizontal component of velocity remains unchanged. Therefore,
$v\cos{30{^\circ} = 10\cos{60{^\circ}}}$
or $v\frac{\sqrt{3}}{2} = \frac{10}{2}\ \ \ or\ \ v = \frac{10}{\sqrt{3}}\ ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.