A projectile is launched from ground level with an initial velocity of 40 m/s at an angle of 30° to the horizontal. Calculate the maximum vertical height reached above ground. (Take g = 10 m/s²)
Answer and explanation
Correct answer: A. 20 m
- Step 1: Vertical component of launch velocity: u_y = u sin 30° = 40 × 0.5 = 20 m/s. Step 2. Maximum height formula: H = u_y² / (2g) = 20² / (2 × 10) = 400 / 20 = 20 m. Answer: 20 m. Why not the others: 40 m uses u² / 2g without angle; 15 m and 30 m represent incorrect trigonometric evaluations.
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