A ball is projected from point $A$ with velocity $20 \mathrm{~m} \mathrm{~s}^{-1}$ at an angle $60^{\circ}$…
A ball is projected from point $A$ with velocity $20 \mathrm{~m} \mathrm{~s}^{-1}$ at an angle $60^{\circ}$ to the horizontal direction. At the highest point $B$ of the path (as shown in figure), the velocity $v \mathrm{~m} \mathrm{~s}^{-1}$ of the ball will be
Zero
$10$
$20$
$10 \sqrt{3}$
Solution
At the highest point of a projectile, vertical component of velocity becomes zero. Therefore at point $B, v=20 \cos 60^{\circ}=20 \times \frac{1}{2}=10 \mathrm{~m} / \mathrm{s}$
