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A ball is projected from point A with velocity 20 m s$^{-1}$ at an angle $60^\circ$ to the horizontal direction. At the highest point B of the path (as shown in figure), the velocity v m s$^{-1}$ of the ball will be: [add image]
A
20
B
$10\sqrt3$
C
Zero
D
10
Detailed Solution
At the topmost point of its trajectory, the particle has only the horizontal component of velocity: $V_{top}=v\cos\theta=20\times\cos60^\circ=20\times\dfrac{1}{2}=10$ m/s.