For what angel of projection the range of projectile equals the maximum height
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Explanation:
Range = u²sin(2θ)/g
Hmax=usinθ/g
=> u²sin(2θ) = usin(θ)
g g
=> 2sin(θ)cos(θ) = sin(θ)
=> sinθ(2cosθ-1)=0
=> sinθ = 0 or cosθ =1/2
=> θ=0 or θ=60°
since angle of projection can't be 0° hence 60° is the required answer....
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