Answer to Question #261569 in Mechanics | Relativity for tamrat

Question

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

Answer 1

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

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

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

$$\dfrac{u^2\sin2\theta}{g} = \dfrac{u^2\sin^2\theta}{2g}\\ \sin2\theta = \dfrac12\sin^2\theta\\ 2\sin\theta\cos\theta = \dfrac12\sin^2\theta\\ 4\cos\theta = \sin\theta\\ \tan\theta = 4\\ \theta =\arctan4 \approx 80\degree$$

Answer. 80 degrees.