A projectile is fired in such a way that its horizontal range is equal to 9.5 times its maximum height. What is the angle of projection?
Horizontal Range = v^{2} sin2θ/g
Height = v^{2} sin^{2}θ/2g
According to the question
v^{2} sin2θ/g= 9.5 x v^{2} sin^{2}θ/2g
=>2v^{2} sinθcosθ/g= 9.5 x v^{2} sin^{2}θ/2g
=> 4/9.5=sinθ/cosθ [On Simpplification]
=>tanθ=4/9.5=> θ==22.83^{0}
Hence the angle of Projection is 22.83^{0} with respect to the horizontal.
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