Derive the formula for the range of a projectile $(R)$. Derive the formula for the maximum range of a projectile.

(N/A) The range $(R)$ is defined as the horizontal distance traveled by a projectile from its initial position $(x=0, y=0)$ to the final position $(x=R, y=0)$ where it returns to the same horizontal level.
At any instant of time $t$,the horizontal position is given by $x = (v_0 \cos \theta_0) t$.
At the time of flight $t_F$,the horizontal distance is $R = (v_0 \cos \theta_0) t_F$.
Since the time of flight is $t_F = \frac{2 v_0 \sin \theta_0}{g}$,we substitute this into the range equation:
$R = (v_0 \cos \theta_0) \left( \frac{2 v_0 \sin \theta_0}{g} \right)$
$R = \frac{v_0^2 (2 \sin \theta_0 \cos \theta_0)}{g}$
Using the trigonometric identity $\sin 2\theta_0 = 2 \sin \theta_0 \cos \theta_0$,we get:
$R = \frac{v_0^2 \sin 2\theta_0}{g}$
For maximum range $(R_{\max})$,the value of $\sin 2\theta_0$ must be maximum,which is $1$.
$\sin 2\theta_0 = 1 \implies 2\theta_0 = 90^\circ \implies \theta_0 = 45^\circ$.
Substituting $\sin 2\theta_0 = 1$ into the range formula:
$R_{\max} = \frac{v_0^2}{g}$.