Projectile Motion

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Maximum height reached by the projectile of KN-003 is:

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5 m

10 m

15 m

20 m

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Subject: physics
Chapter: kinematics
Topic: projectile motion
Difficulty: Medium
Year: 2025

Solution

Correct Answer:

5 m

From KN-003, $u = 20 m/s$ and $θ = 3 0^{\circ}$ . The formula for maximum height $H_{ma x}$ is: $H_{ma x} = \frac{u ^{2} s i n ^{2} θ}{2 g}$ Substituting values:

$sin 3 0^{\circ} = 0.5 ⟹ sin^{2} 3 0^{\circ} = 0.25$
$u^{2} = 400$
$2 g = 20$ $H_{ma x} = \frac{400 \times 0.25}{20} = \frac{100}{20} = 5 m$ .

This medium difficulty physics question is from the chapter kinematics, covering the topic of projectile motion. It appeared in the 2025 exam.

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Projectile Motion

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