Projectile Motion

Any object in flight after being launched with some initial velocity, moving thereafter only under the influence of Earth's gravity, is called a projectile, and its motion is projectile motion. Everyday examples abound: a stone thrown at a fruit tree, a cricket ball bowled toward a batsman, a basketball arcing toward the hoop.

A projectile's velocity has two components — horizontal (along x) and vertical (along y) — and because gravity acts purely vertically downward, these two components behave completely differently. The horizontal component stays unchanged throughout the flight, since no force acts along that direction; the vertical component changes uniformly, governed by an acceleration $a_y = -g$ (taking upward as positive), while $a_x = 0$. Unless stated otherwise, air resistance and other retarding forces are neglected.

Suppose the projectile is launched from the ground (the origin) with initial speed $u$ at angle $\theta$ above the horizontal. Its initial velocity components are $u_x = u\cos\theta$ and $u_y = u\sin\theta$. Applying the plane equations of motion (section 3.3.3) with $a_x=0$ and $a_y=-g$:

Velocity components at time $t$: $v_x = u\cos\theta$ (Eq. 3.38, unchanged for all t) and $v_y = u\sin\theta - gt$ (Eq. 3.39). Since $0 < \theta < 90°$, the vertical component starts out positive (upward).

Displacement components at time $t$: $s_x = (u\cos\theta)\,t$ (Eq. 3.40) and $s_y = (u\sin\theta)\,t - \tfrac{1}{2}gt^2$ (Eq. 3.41). The instantaneous direction of motion at any time $t$ makes an angle $\alpha$ with the horizontal given by $\tan\alpha = v_y(t)/v_x(t)$ (Eq. 3.42).

As the projectile rises, its vertical velocity steadily shrinks, reaching exactly zero at the point of maximum height — after which it starts increasing again but in the downward sense, with the horizontal velocity component staying constant throughout. By the trajectory's symmetry, the time taken to rise equals the time taken to fall back to the same height (in the idealised no-air-resistance case). Setting $v_y=0$ at $t=t_0$ in Eq. (3.39): $0 = u\sin\theta - gt_0$, so $t_0 = \dfrac{u\sin\theta}{g}$ (Eq. 3.43) is the time to reach the top, and the total time of flight is $T = 2t_0$.

The horizontal range R (the total horizontal distance covered before landing) follows from substituting $t=T$ into Eq. (3.40): $R = u\cos\theta \cdot T = u\cos\theta \cdot \dfrac{2u\sin\theta}{g} = \dfrac{2u_x u_y}{g} = \dfrac{u^2\sin 2\theta}{g}$ (Eq. 3.44), using the identity $2\sin\theta\cos\theta = \sin2\theta$. Since $\sin2\theta$ is largest (equal to 1) when $2\theta = 90°$, i.e. $\theta = 45°$, the range is maximum at a launch angle of 45°, giving $R_{max} = \dfrac{u^2}{g}$.

The maximum height H is the vertical displacement at $t=t_0$, from Eq. (3.41): $H = u\sin\theta\cdot t_0 - \tfrac{1}{2}g t_0^2 = \dfrac{u^2\sin^2\theta}{2g} = \dfrac{u_y^2}{2g}$ (Eq. 3.45).

These formulas for T, R, $R_{max}$ and H all assume gravity is the only force acting; in reality air resistance makes the time of ascent shorter than the time of descent, and to actually maximise the range the launch angle needs to be a little more than 45°, with the true maximum range falling short of the idealised $u^2/g$. …