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

Range is defined as the horizontal distance travelled by a projectile from its initial position $y=0$ ) to the final position (where it passes $y=0$ during its fall) is called the Range of the projectile.

or it is the distance travelled from its initial position $(x=0, y=0)$ to final position $(x=\mathrm{R}, y=0)$ is called 'R'.

At any instant of time keeping $x=\mathrm{R}$ and $\mathrm{t}=\mathrm{t}_{\mathrm{F}}$ then $x=\left(v_{0} \cos \theta_{0}\right) t$

$\therefore \mathrm{R}=\left(v_{\mathrm{o}} \cos \theta_{\mathrm{o}}\right) t_{\mathrm{F}}$

$\therefore \mathrm{R}=\left(v_{\mathrm{o}} \cos \theta_{\mathrm{o}}\right)\left(\frac{2 v_{o} \sin \theta_{o}}{g}\right)\left[\because t_{\mathrm{F}}=\frac{2 v_{o} \sin \theta_{o}}{g}\right]$

$\therefore \mathrm{R}=\frac{2 v_{o}^{2} \sin \theta_{o} \cos \theta_{o}}{g}$

$\therefore \mathrm{R}=\frac{v_{o}^{2}\left(\sin \theta_{o} \cos \theta_{o}\right)}{g}$

$\therefore \mathrm{R}=\frac{v_{o}^{2} \sin 2 \theta_{o}}{g}$

$\mathrm{R}_{\max }$ is given by $\frac{v_{o}^{2}}{g}$

When $\sin 2 \theta_{0}=1$

For $\mathrm{R}_{\max } \quad \sin 2 \theta_{\mathrm{o}}=1 .$

$\therefore 2 \theta_{\mathrm{o}}=90^{\circ}$

$\therefore \theta_{\mathrm{o}}=45^{\circ}$

If the object is at an angle $\theta_{o}=45^{\circ}$, then its range is maximum. The value of $R_{\max }$ depends upon its initial velocity.