Projectile motion

Example

A particle of mass $m$ moves in the vertical plane with coordinates $(x, z)$, under uniform gravity $(0, -mg)$ with $g > 0$ and without air resistance. It starts at the origin at time $0$ with speed $v > 0$ at the angle $\theta \in (0, \pi/2)$ above the horizontal. Newton's second law gives $\ddot x = 0$ and $\ddot z = -g$, so $$x(t) = v t \cos\theta, \qquad z(t) = v t \sin\theta - \tfrac12 g t^2.$$ Check: $m(\ddot x, \ddot z) = (0, -mg)$, and $(x(0), z(0)) = (0, 0)$, $(\dot x(0), \dot z(0)) = (v\cos\theta, v\sin\theta)$. Eliminating $t = x/(v \cos\theta)$ gives the path $$z = x \tan\theta - \frac{g}{2 v^2 \cos^2\theta}\, x^2,$$ a parabola. The particle returns to $z = 0$ at time $2 v \sin\theta / g$, at the distance $R = v^2 \sin(2\theta)/g$. The range $R$ is largest for $\theta = \pi/4$, and the greatest height is $v^2 \sin^2\theta/(2g)$.