Q. A ball is projected with a velocity, 10 $ms^{-1}$, at an angle of 60° with the vertical direction. Its speed at the highest point of its trajectory will be :
A) 10 $ms^{-1}$
B) Zero
C) $5\sqrt{3}\ ms^{-1}$
D) 5 $ms^{-1}$
Explanation
[Solution] At the highest point only the horizontal component remains: $10\sin 60^\circ = 5\sqrt{3}\ ms^{-1}$. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.