Projectile Motion

Projectile motion is basically a vertical uniform acceleration motion (with a = 9.81 m s^-2 downward) combined with a horizontal constant speed motion.
(xmdemo)
(xmdemo)

$\begin{array}{l}{{u}_{x}}=u\cos \theta \\{{u}_{y}}=u\sin \theta \end{array}$

${{v}_{y}}={{u}_{y}}-gt$
${{s}_{y}}={{u}_{y}}t-\frac{1}{2}g{{t}^{2}}$
${{v}_{y}}^{2}={{u}_{y}}^{2}-2g{{s}_{y}}$

${{v}_{x}}={{u}_{x}}$
${{s}_{x}}={{u}_{x}}t$

projectile

$\begin{aligned}({{v}^{2}}&={{u}^{2}}+2as)\\0&={{u}_{y}}^{2}-2gH\\H&=\frac{{{u}_{y}}^{2}}{2g}\end{aligned}$

The time taken to reach the peak t_p, and the time of flight t_f are both dependent only on u_y. In fact, both are proportional to u_y.

$\begin{aligned}(v&=u+at)\\0&={{u}_{y}}-g{{t}_{p}}\\{{t}_{p}}&=\frac{{{u}_{y}}}{g}\\{{t}_{f}}&=2{{t}_{p}}=\frac{2{{u}_{y}}}{g}\end{aligned}$

$\begin{aligned}R={{u}_{x}}{{t}_{f}}={{u}_{x}}\frac{2{{u}_{y}}}{g}=\frac{2{{u}_{x}}{{u}_{y}}}{g}\end{aligned}$

$\begin{aligned}R&=\frac{2{{u}_{x}}{{u}_{y}}}{g}\\&=\frac{2(u\cos \theta )(u\sin \theta )}{g}\\&=\frac{{{u}^{2}}\sin 2\theta }{g}\end{aligned}$

Since sin2θ is symmetrical about θ = 45°, projectiles launched at θ and (90° – θ) land at the same spot.

Click here for illustrations of how various parameters affect a projectile’s flight paths.

xmPhysics