Post-UTME Physics 2009 Objective — Question 7
Question 7 of 10 from the Post-UTME Screening (Post-UTME) Physics 2009 Objective paper, with the correct answer and a full explanation.
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An object is projected with a velocity of 100ms⁻¹ from the ground level at an angle to the vertical. Total time of flight of the projectile is 10s, calculate the angle (g=10ms⁻²)
- A. 0°
- B. 30°
- C. 45°
- D. 60°Correct
Explanation
T=2u cosθ/g (θ from vertical) gives 10=2(100)cosθ/10, so cosθ=0.5, θ=60°.
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