Projectile Motion

Independence of horizontal and vertical components, free fall, projection at an angle, and the effects of air resistance on trajectories.

Spec Points Covered
  • State that the horizontal and vertical components of projectileAn object moving freely under gravity after being launched. Horizontal and vertical components of motion are independent. motion are independent of each other.
  • Resolve the initial velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹. of a projectileAn object moving freely under gravity after being launched. Horizontal and vertical components of motion are independent. into horizontal and vertical components.
  • Apply SUVAT equations separately to the horizontal and vertical motion.
  • Calculate time of flight, maximum height and range for projectiles.
  • Describe the effects of air resistanceThe opposition to currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A). flow. The ratio of potential difference to currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A).. Measured in ohms (Ω). on the trajectory of a projectileAn object moving freely under gravity after being launched. Horizontal and vertical components of motion are independent..
Notes
01 A projectile is any object moving freely under gravity in two dimensions Projectile 3.4.1.4 → 02 Horizontal and vertical components of projectile motion are independent $u_x = u \cos \theta$ 3.4.1.4 → 03 Time of flight is found from the vertical motion $t_{\text{max}} = \frac{u \sin \theta}{g}$ 3.4.1.4 → 04 Maximum height uses \(v^{2}\) = \(u^{2}\) + 2as with v = 0 $H = \frac{(u \sin \theta)^2}{2g}$ 3.4.1.4 → 05 Range is found from horizontal velocity multiplied by total time $R = \frac{u^2 \sin 2\theta}{g}$ 3.4.1.4 → 06 Horizontal projection: initial vertical velocity is zero 3.4.1.4 → 07 Air resistance reduces range and maximum height of a projectile 3.4.1.4 →
Σ Key Equations Full Reference →
On Data Sheet
Not on Data Sheet
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.
Q Retrieval Practice All 10 Questions →
Q1. What is meant by the independence of horizontal and vertical components in projectile motion?
  • The horizontal and vertical motions are independent: gravity only affects the vertical component, and horizontal velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹. stays constant (no air resistanceThe opposition to currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A). flow. The ratio of potential difference to current. Measured in ohms (Ω).).
  • They are linked only by time.
Q2. How do you resolve the initial velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹. of a projectile launched at angle theta?
  • Horizontal: u cos(theta).
  • Vertical: u sin(theta).
Q3. What is the vertical velocity at the maximum height of a projectile?
Zero.
Q4. How do you find the time of flight for a symmetric projectile?
Time of flight = 2u sin(theta) / g.
Q5. Write the equation for the maximum height of a projectile.
H = (u sin theta)^2 / (2g).