A stone is thrown with a speed of 10ms at an angle of 30° to the horizontal, Calculate its time of flight, (g = 10ms2)

Analyze the question: A projectile is launched at 10 m/s at 30° above the horizontal. We need the total time it stays in the air (time of flight). The speed is taken as 10 m/s and g = 10 m/s².

Concept: For a projectile launched and landing at the same level, only the vertical component of velocity determines the time of flight:

T = (2u sinθ) / g

This works because the stone rises with vertical velocity u sinθ, stops at the top after u sinθ / g seconds, and takes the same time to fall back.

Solution:

1. Vertical component of velocity: u sinθ = 10 × sin 30° = 10 × 0.5 = 5 m/s

2. Substitute into the formula: T = (2 × 5) / 10 = 10 / 10 = 1.0 s

Check: Time to the top = 5/10 = 0.5 s. The fall takes another 0.5 s, so the total is 1.0 s.

Answer: B. 1.0 s

Exam tip: Remember that sin 30° = 0.5. When u = 10 and g = 10, the formula simplifies to T = 2 × 10 × sinθ / 10 = 2 sinθ. So at 30° the time of flight is just 2 × 0.5 = 1 s. Also, options like 1.7 s (which would come from using cos 30°, since 2 × 10 × 0.866 / 10 ≈ 1.7) are distractors. Always use sin for time of flight and maximum height, and cos for the horizontal range component.