Projectile Motion

Horizontal and vertical motions are independent.

• Horizontal: No acceleration (ignoring air resistance) → constant velocity

  • $a_x = 0$
  • $s_x = u_x t$

• Vertical: Acceleration due to gravity ($g = 9.81$ m/s² downwards)

  • $a_y = -g$ (if up is positive)
  • Apply SUVAT equations

The time $t$ is the same for both components — this links them.

For a projectile launched at speed $u$ at angle $\theta$ to the horizontal:

$u_x = u\cos\theta \qquad u_y = u\sin\theta$

Time of Flight (return to launch height)

At landing, $s_y = 0$: $0 = u\sin\theta \cdot t - \frac{1}{2}gt^2$ $t(u\sin\theta - \frac{1}{2}gt) = 0$ $T = \frac{2u\sin\theta}{g}$

Maximum Height

At the top, $v_y = 0$: $v_y^2 = u_y^2 - 2gs_y$ $0 = u^2\sin^2\theta - 2gH$ $H = \frac{u^2\sin^2\theta}{2g}$

Range (horizontal distance)

$R = u_x \times T = u\cos\theta \times \frac{2u\sin\theta}{g} = \frac{u^2\sin2\theta}{g}$

Maximum range occurs at $\theta = 45°$ (since $\sin90° = 1$).

Equation of Trajectory

Eliminating $t$: $y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}$

This is a parabola.

If launched horizontally from height $h$:

• $u_x = u$, $u_y = 0$
• Time to fall: $h = \frac{1}{2}gt^2$ → $t = \sqrt{2h/g}$
• Range: $R = u \times t = u\sqrt{2h/g}$
• Impact speed: $v = \sqrt{u^2 + (gt)^2}$

• Horizontal: constant velocity ($a_x = 0$)
• Vertical: constant acceleration ($a_y = g$)
• Resolve launch velocity: $u_x = u\cos\theta$, $u_y = u\sin\theta$
• Time is the common variable linking both components
• $T = 2u\sin\theta/g$, $H = u^2\sin^2\theta/2g$, $R = u^2\sin2\theta/g$
• Max range at 45°; trajectory is a parabola