Q104 / 200 · NEET 2023 (Manipur) · Physics

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 Figure

  • A
    Zero
  • Correct answer
    B
    $10$
  • C
    $20$
  • D
    $10 \sqrt{3}$

Solution

At the highest point of a projectile, vertical component of velocity becomes zero. Therefore at point $B, v=20 \cos 60^{\circ}=20 \times \frac{1}{2}=10 \mathrm{~m} / \mathrm{s}$

Hence, the correct answer is option: 'B'.

Relevant Questions