Q 11-03-032NEETJEE MainMedium
The horizontal range of a projectile is $4\sqrt{3}$ times its maximum height. The angle of projection is
Answer: (A) $30°$
Concept. $\dfrac{R}{H} = 4\cot\theta$.
Step 1. Solve. $4\cot\theta = 4\sqrt3 \Rightarrow \theta = 30°$.
Answer. $30°$ (Option A).
Solution by Sreeraj P, M.Sc Physics