Question

# A projectile is fired in such a way that its horizontal range is equal to 9.5...

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 = v2 sin2θ/g

Height = v2 sin2θ/2g

According to the question

v2 sin2θ/g= 9.5 x v2 sin2θ/2g

=>2v2 sinθcosθ/g= 9.5 x v2 sin2θ/2g

=> 4/9.5=sinθ/cosθ [On Simpplification]

=>tanθ=4/9.5=> θ= =22.830

Hence the angle of Projection is 22.830 with respect to the horizontal.

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