A-Level — mathsAdvanced3 min read

Projectile Motion

Analyse projectile motion at A-Level. Resolve velocity into horizontal and vertical components, find range, max height, and time of flight.

Tutor AI Team
Tutor AI Team

Projectile motion occurs when an object is launched into the air and moves under gravity alone (ignoring air resistance). The horizontal and vertical components of motion are analysed independently.

Core Principles

Independence of Horizontal and Vertical Motion

  • Horizontal: constant velocity (no acceleration).

    • x=ucos⁡θ⋅tx = u\cos\theta \cdot t

  • Vertical: constant acceleration gg downward.

    • vy=usin⁡θ−gtv_y = u\sin\theta - gt
    • y=usin⁡θ⋅t−12gt2y = u\sin\theta \cdot t - \frac{1}{2}gt^2
    • vy2=(usin⁡θ)2−2gyv_y^2 = (u\sin\theta)^2 - 2gy

Key Quantities

Time of Flight (returns to same level)

T=2usin⁡θgT = \frac{2u\sin\theta}{g}

Maximum Height

H=u2sin⁡2θ2gH = \frac{u^2\sin^2\theta}{2g}

Range (horizontal distance)

R=u2sin⁡2θgR = \frac{u^2\sin 2\theta}{g}

Maximum range when θ=45°\theta = 45°.

Worked Example: Ball Kicked at an Angle

Problem

A ball is kicked at 20 m/s at 30° above the horizontal. Find:

Horizontal component: ux=20cos⁡30°=17.32u_x = 20\cos 30° = 17.32 m/s. Vertical component: uy=20sin⁡30°=10u_y = 20\sin 30° = 10 m/s.

Time of flight: T=2×109.8=2.04T = \frac{2 \times 10}{9.8} = 2.04 s.

Max height: H=10019.6=5.10H = \frac{100}{19.6} = 5.10 m.

Range: R=17.32×2.04=35.3R = 17.32 \times 2.04 = 35.3 m.

Solution

Worked Example: Launched from a Height

Problem

A stone is thrown horizontally at 15 m/s from a cliff 45 m high.

ux=15u_x = 15, uy=0u_y = 0.

Time to fall: 45=12(9.8)t245 = \frac{1}{2}(9.8)t^2 → t=3.03t = 3.03 s.

Horizontal distance: 15×3.03=45.515 \times 3.03 = 45.5 m.

Speed on impact: vx=15v_x = 15, vy=9.8×3.03=29.7v_y = 9.8 \times 3.03 = 29.7. v=152+29.72=33.3v = \sqrt{15^2 + 29.7^2} = 33.3 m/s.

Solution

Worked Example: Finding the Angle

Problem

A ball is launched at 25 m/s and reaches max height 10 m. Find the launch angle.

H=u2sin⁡2θ2gH = \frac{u^2\sin^2\theta}{2g} → 10=625sin⁡2θ19.610 = \frac{625\sin^2\theta}{19.6} → sin⁡2θ=0.3136\sin^2\theta = 0.3136 → sin⁡θ=0.56\sin\theta = 0.56 → θ=34.1°\theta = 34.1°.

Solution

The Equation of the Trajectory

Eliminate tt from parametric equations:

y=xtan⁡θ−gx22u2cos⁡2θy = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}

This is a parabola.

Strategy

  1. Resolve the initial velocity into horizontal and vertical components.
  2. Horizontal: use x=uxtx = u_x t (constant velocity).
  3. Vertical: use SUVAT with a=−ga = -g.
  4. Link horizontal and vertical using time tt.

Common Mistakes

  • Forgetting to resolve the initial velocity into components.
  • Mixing up horizontal and vertical. Horizontal: no acceleration. Vertical: acceleration = gg.
  • Sign errors with gg. If upward is positive, a=−ga = -g.
  • Using range/time formulas when not returning to the same level. The standard formulas assume the projectile lands at the same height it was launched.

Practice Problems

  1. Problem 1

    A ball is thrown at 30 m/s at 60° above the horizontal. Find the time of flight, max height, and range.

    Problem 2

    A stone is projected horizontally at 10 m/s from the top of a 20 m building. Find the time to reach the ground and the distance from the base.

    Problem 3

    A projectile has range 100 m when launched at 45°. Find the initial speed.

    Problem 4

    Show that the maximum range for a given speed occurs at θ=45°\theta = 45°.

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Key Takeaways

  • ✓

    Horizontal: constant velocity, x=ucos⁡θ⋅tx = u\cos\theta \cdot t.

  • ✓

    Vertical: SUVAT with a=ga = g downward.

  • ✓

    Time of flight =2usin⁡θg= \frac{2u\sin\theta}{g} (same level).

  • ✓

    Max height =u2sin⁡2θ2g= \frac{u^2\sin^2\theta}{2g}.

  • ✓

    Range =u2sin⁡2θg= \frac{u^2\sin 2\theta}{g}. Maximum at θ=45°\theta = 45°.

  • ✓

    The trajectory is a parabola.

Tags:
#a-level#maths#mechanics#projectiles#kinematics

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