At what angle of projection of a projectile the range becomes half of its maximum value?
Correct answer: A. 15
- A. 15
- B. 20
- C. 30
- D. 40
Explanation
R=Vi^2 sin 2θ/g At given condition: Maximum range=Rmax=vi^2θ/g For maximum range, sin2θ=1 which is possible if we put angle as 45 degrees but to get half of the maximum range we should be putting θ so as to get sin2θ as 0.5. It is known that sin 30=0.5 so if we put θ as 15 we will get sin2(15)=sin 30=0.5.
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