# 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.

## Solved Problems from IIT JEE