\[ u_{x} = u \cos \theta \] \[ u_{y} = u \sin \theta \] \[ s_x = v_{x} t = (u \cos \theta)t \] \[ s_y = v_y t - \frac{1}{2} g t^2 = (u \sin \theta ) t - \frac{1}{2} g t^2 \] \[ (u \sin \theta )t - \frac{1}{2} g t^2 = 0 \] \[ t (u \sin \theta - \frac{1}{2} g t) = 0 \] \[ t = 0 \text{ or } t = \frac{2 u \sin \theta}{g}\] Two answers for when the projectile is at "ground level" - at the beginning and end of flight. We choose the latter. \[ T = \text{time of flight} = \frac{2 u \sin \theta}{g} \] \[ R = \text{range} = u_{x} T = (u \cos \theta) \frac{2 u \sin \theta}{g} \] \[ R = \frac{2 u^2 \sin \theta \cos \theta}{g} \] Using the trig identity, we have: $$2 \sin \theta \cos \theta = \sin (2\theta)$$ \[ R = \frac{u^2 \sin 2\theta}{g} \] Maximum R will occur when the sin function is a maximum - i.e. when it reaches an output of 1 when its input is 90\(^\circ\) \[ (\sin2\theta)_{max} = 1 \quad\rightarrow\quad 2\theta = 90^\circ \quad\rightarrow\quad \theta = 45^\circ\] \[ R_{\text{max}} = \frac{u^2 \cdot (1)}{g} = \frac{u^2}{g} \] \[ \boxed{R_{\text{max}} = \frac{u^2}{g}}\]