# Projectiles

## Concepts and Formulas Let a particle be projected from a point O with an initial velocity $u$ at an angle $\theta$ with the horizontal. The motion lies in a vertical plane containing the point O and the initial velocity vector. Let $x$ be the horizontal and $y$ be the vertical directions in this plane. If we take O as the origin $(x=0,y=0)$ and the instant of projection as time $t=0$ then initial position, velocity, and acceleration of the particle are given by, \begin{align} x&=0, && u_x&=u\cos\theta, && a_x&=0; \\ y&=0, && u_y&=u\sin\theta, && a_y&=-g; \end{align} where $g$ is the acceleration due to gravity. Use equation of motion, $s=ut+\frac{1}{2}at^2$, to get the position at the time $t$ as, \begin{align} x=ut\cos\theta,&& y=ut\sin\theta-\tfrac{1}{2}gt^2. \end{align} Eliminate $t$ from the expressions of $x$ and $y$ to get the equation of trajectory, \begin{align} y=x\tan\theta-\frac{g}{2u^2\cos^2\theta} x^2, && \text{(equation of a parabola)}. \end{align}

The time of flight is given by \begin{align} T=\frac{2u\sin\theta}{g}, \end{align}

and the range is \begin{align} R=\frac{u^2\sin2\theta}{g}. \end{align} The maximum range for given initial velocity is $R_\text{max}=u^2/g$ and it occurs at $\theta=45^\mathrm{o}$.

The maximum height of the projectile is given by \begin{align} H=\frac{u^2 \sin^2 \theta}{2g}. \end{align} Note that $v_y=0$ at maximum height.