Question #261569

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
2021-11-07T19:28:42-0500

The range RR and maximum height hh are given as follows (see https://courses.lumenlearning.com/boundless-physics/chapter/projectile-motion/):


R=u2sin2θgh=u2sin2θ2gR = \dfrac{u^2\sin2\theta}{g}\\ h = \dfrac{u^2\sin^2\theta}{2g}

Equating these expressions, find the angle of projection θ\theta:


u2sin2θg=u2sin2θ2gsin2θ=12sin2θ2sinθcosθ=12sin2θ4cosθ=sinθtanθ=4θ=arctan480°\dfrac{u^2\sin2\theta}{g} = \dfrac{u^2\sin^2\theta}{2g}\\ \sin2\theta = \dfrac12\sin^2\theta\\ 2\sin\theta\cos\theta = \dfrac12\sin^2\theta\\ 4\cos\theta = \sin\theta\\ \tan\theta = 4\\ \theta =\arctan4 \approx 80\degree

Answer. 80 degrees.


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