AIPMT PRELIMS 2012 Physics Projectile Motion MCQ Question
The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is
θ = 45°
θ = tan⁻¹(1/4)
θ = tan⁻¹(4)
θ = tan⁻¹(2)
Correct Answer
Detailed Explanation
For the horizontal range and maximum height to be equal, tan θ must equal 4. Solving for θ gives θ = tan⁻¹(4).
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