## Essential University Physics: Volume 1 (4th Edition)

We know that $x=\frac{v_{\circ}^2}{g} sin (2\theta)$ After differentiating, we obtain: $x^{\prime}=(\frac{v_{\circ}^2}{g})(-2cos(2\theta))$ as $x^{\prime}=0$ $\implies cos(2\theta)=0$ $2\theta=90$ $\theta=45^{\circ}$