Key Equations

Projectile Motion — AQA A-Level Physics

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Horizontal component of velocity
$$u_x = u \cos \theta$$
  • Where:
    • $u_x$ = horizontal velocity (m s⁻¹)
    • $u$ = launch speed (m s⁻¹)
    • $θ$ = angle to horizontal
Constant throughout the flight (no horizontal acceleration).
Vertical component of velocity
$$u_y = u \sin \theta$$
  • Where:
    • $u_y$ = initial vertical velocity (m s⁻¹)
    • $u$ = launch speed (m s⁻¹)
    • $θ$ = angle to horizontal
Changes throughout flight due to gravitational acceleration.
Time to maximum height
$$t_{\text{max}} = \frac{u \sin \theta}{g}$$
  • Where:
    • $t_max$ = time to max height (s)
    • $u$ = launch speed (m s⁻¹)
    • $θ$ = angle to horizontal
    • $g$ = 9.81 m s⁻²
Total time of flight = 2 times this value (symmetric trajectory only).
Maximum height
$$H = \frac{(u \sin \theta)^2}{2g}$$
  • Where:
    • $H$ = maximum height (m)
    • $u$ = launch speed (m s⁻¹)
    • $θ$ = angle to horizontal
    • $g$ = 9.81 m s⁻²
Derived from \(v^{2}\) = \(u^{2}\) + 2as with v = 0.
Range (symmetric trajectory)
$$R = \frac{u^2 \sin 2\theta}{g}$$
  • Where:
    • $R$ = range (m)
    • $u$ = launch speed (m s⁻¹)
    • $θ$ = angle to horizontal
    • $g$ = 9.81 m s⁻²
Only valid when launch and landing heights are the same.
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