Projectile Motion

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A projectile is fired at an angle of 30° with the horizontal with an initial velocity of 40 m/s. What is the maximum height reached? (g = 10 m/s²)

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20 m

40 m

60 m

80 m

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

Solution

Correct Answer:

20 m

Maximum Height Formula: $H = \frac{u ^{2} s i n ^{2} θ}{2 g}$ .
Identify Variables:
- $u = 40$ m/s.
- $θ = 3 0^{\circ}$ .
- $g = 10$ m/s $^{2}$ .
Calculation:
- $sin 3 0^{\circ} = 0.5$ .
- $H = \frac{( 40 ) ^{2} \times ( 0.5 ) ^{2}}{2 \times 10} = \frac{1600 \times 0.25}{20}$ .
Final Value: $H = \frac{400}{20} = 20$ m.
Note: The maximum height depends only on the vertical component of the initial velocity ( $u sin θ$ ).

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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