Question #261570

A projectile is fired in such a way that its horizontal range is equal to its maximum height. What is the angle of projection?


1
Expert's answer
2021-11-07T19:28:37-0500
R=3HR=3Hv2sin2θg=2v2sinθcosθg=3v2sin2θ2g\frac{v^2\sin{2\theta}}{g}=2\frac{v^2\sin{\theta}\cos{\theta}}{g}=3 \frac{v^2\sin^2{\theta}}{2g}tanθ=43\tan \theta=\frac{4}{3}θ=53°\theta=53\degree

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