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?
Homework Answers
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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