considering the rock; d=ymaxd=y_{max}d=ymax
v2sin2θg=v2sin2θ2g\frac{v^2sin2\theta}{g}=\frac{v^2sin^2\theta}{2g}gv2sin2θ=2gv2sin2θ
2sinθcosθ=sin2θ22sin\theta cos\theta=\frac{sin^2\theta}{2}2sinθcosθ=2sin2θ
2cosθ=sinθ22cos\theta=\frac{sin\theta}{2}2cosθ=2sinθ
tanθ=4tan\theta=4tanθ=4
θ=tan−14=75.960\theta=tan^{-1}4=75.96^0θ=tan−14=75.960
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