1. 10\ ms⁻¹ 2. Zero 3. 5√(3)\ ms⁻¹ 4. 5\ ms⁻¹

Answer: 3. 5√(3)\ ms⁻¹ The angle is measured from the vertical, so the angle with the horizontal is 30^° . At the highest point only the horizontal component remains: v=10cos30^°=10×sqrt3/2=5sqrt3\ m/s . Option (3).

Topic: Projectile motion· Kinematics

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 :

1. \(10\ \text{ms}^{-1}\)

2. Zero

3. \(5\sqrt{3}\ \text{ms}^{-1}\)

4. \(5\ \text{ms}^{-1}\)

Doubtly Asked 30 Oct 2026 NEET UG 2022 Practice test neet Save WhatsApp Report

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Answer: 3. \(5\sqrt{3}\ \text{ms}^{-1}\)

The angle is measured from the vertical, so the angle with the horizontal is \(30^\circ\).

At the highest point only the horizontal component remains: \(v=10\cos30^\circ=10\times\dfrac{\sqrt3}{2}=5\sqrt3\ \text{m/s}\).

Option (3).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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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 :

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