A ball is projected from point A with velocity 20 m s⁻¹ at an angle 60° to the horizontal direction. At the highest point B of the path (as shown in figure), the velocity v ms⁻¹ of the ball will be: (a) 20 (b) 10√(3) (c) Zero (d) 10

Answer: (d). At the highest point the vertical velocity is zero; only the horizontal component ucos θ remains. v = 20cos 60° = 20 × 1/2 = 10 m s⁻¹ So the answer is (d) 10.

A ball is projected from point A with velocity 20 m s⁻¹ at an angle… | NEET 2023

Q9NEET 2023Easy

Velocity, Momentum & Energy in Flight

A ball is projected from point A with velocity $20\text{ m s}^{-1}$ at an angle $60^\circ$ to the horizontal direction. At the highest point $\text{B}$ of the path (as shown in figure), the velocity $v\text{ ms}^{-1}$ of the ball will be:

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Answer: (d)

At the highest point the vertical velocity is zero; only the horizontal component $u\cos\theta$ remains.
$v = 20\cos 60^\circ = 20 \times \frac{1}{2} = 10\text{ m s}^{-1}$
So the answer is (d) 10.

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