A projectile is fired with an initial velocity u at an angle theta with the horizontal. At what angle of projection is the horizontal range equal to…

The horizontal range is $R = \frac{u^2 \sin 2\theta}{g}$ and maximum height is $H = \frac{u^2 \sin^2 \theta}{2g}$. Given $R = 4H$, so $\frac{u^2 \sin 2\theta}{g} = 4 \left(\frac{u^2 \sin^2 \theta}{2g}\right) \implies 2\sin\theta\cos\theta = 2\sin^2\theta \implies \tan\theta = 1 \implies \theta = 45^\circ$.