Q. A bullet is fired with a velocity $u$ making an angle of $60{^\circ}$ with the horizontal plane. The horizontal component of the velocity of the bullet when it reaches the maximum height is
A) $u$
B) $0$
C) $\frac{\sqrt{3u}}{2}$
D) $u/2$
Explanation
(d) At maximum height $H$, the horizontal component of the velocity of the bullet $= u\cos\theta = u\cos{60{^\circ}} = u/2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.