NECO 2002 Physics · Question 23

An object is projected with a velocity of 50 ms^ -1 from ground level at an angle θ to the vertical. If the…

Question

An object is projected with a velocity of 50 $ms^{-1}$ from ground level at an angle θ to the vertical. If the total time of flight of the projectile is 5s, what is the value of θ? (g = 10 $ms^{-2}$ )

A.0°
B.30°
C.45°
D.60°
E.90°

Answer: D. 60°

Explanation written with AI and checked against the answer key

Analyze the question: A projectile is launched at 50 m/s from ground level, and the angle θ is measured from the vertical (not the horizontal, which is the usual convention). The time of flight is 5 s. Concept: For a projectile landing at the same level it was launched from: T = 2u cos θ / g (when θ is measured from the vertical) The vertical component of velocity is u cos θ when θ is from the vertical. (If θ were from the horizontal, it would be u sin θ.) Solution: 5 = 2(50)cos θ / 10 5 = 10 cos θ cos θ = 0.5 θ = 60° Check with the horizontal convention: If θ were from the horizontal, 5 = 10 sin θ gives sin θ = 0.5, so θ = 30°, which is option B. But the question clearly states the angle is to the vertical, so the vertical component is u cos θ and the answer is 60°. Verification: The vertical component is 50 cos 60° = 25 m/s. Time of flight = 2(25)/10 = 5 s ✓. Answer: D (60°) Tip: Always check what the angle is measured from. Angle from the horizontal means the vertical component is u sin θ. Angle from the vertical means the vertical component is u cos θ. Examiners often use this wording to catch students who rush.

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