Projectiles

Projectiles

A projectile is an object moving freely under gravity in two dimensions. At A-Level, you model projectile motion by treating the horizontal and vertical components independently.

Key Assumptions

The standard projectile model assumes:

• The projectile is a particle (no air resistance, no spin)
• Gravity is the only force, acting vertically downward at g = 9.8 m/s²
• The ground is flat and horizontal (no curvature of the Earth)
• Horizontal velocity is constant (no air resistance)
• Vertical motion follows SUVAT with a = -g (taking upward as positive)

Resolving the Initial Velocity

If a projectile is launched at speed u at angle θ to the horizontal:

• Horizontal component: uₓ = ucosθ (constant throughout the motion)
• Vertical component: u_y = usinθ (changes due to gravity)

The Equations of Motion

Horizontal: Since there is no horizontal acceleration:

• x = uₓt = (ucosθ)t

Vertical: Using SUVAT with initial vertical velocity usinθ and acceleration -g:

• v_y = usinθ - gt
• y = (usinθ)t - ½gt²
• v_y² = (usinθ)² - 2gy

Key Results

Time to reach maximum height: At the highest point, v_y = 0.

0 = usinθ - gt, so t = usinθ / g

Maximum height: Substituting into the vertical displacement formula:

H = u²sin²θ / (2g)

Total time of flight (returning to launch height):

T = 2usinθ / g (double the time to maximum height, by symmetry)

Range (horizontal distance at landing):

R = u²sin2θ / g

The range is maximised when θ = 45° (since sin2θ is maximised when 2θ = 90°).

Worked Example 1: Basic Projectile

A ball is kicked at 20 m/s at 30° above the horizontal from ground level. Find the maximum height, time of flight, and range.

uₓ = 20cos30° = 17.32 m/s, u_y = 20sin30° = 10 m/s

Maximum height: H = 10² / (2 × 9.8) = 100/19.6 = 5.10 m

Time of flight: T = 2(10)/9.8 = 20/9.8 = 2.04 s

Range: R = 17.32 × 2.04 = 35.3 m

Worked Example 2: Projectile from a Height

A stone is thrown horizontally at 15 m/s from the top of a cliff 40 m high. Find how far from the base of the cliff it lands.

Horizontal: uₓ = 15 m/s.

Vertical (taking downward as positive, starting from rest vertically): u_y = 0, a = 9.8.

Find time to fall 40 m: 40 = ½(9.8)t², so t² = 80/9.8 = 8.163, t = 2.857 s

Horizontal distance: x = 15 × 2.857 = 42.9 m

Worked Example 3: Finding the Angle of Impact

Continuing the cliff example, find the speed and direction when the stone hits the ground.

v_y = 0 + 9.8(2.857) = 28.0 m/s (downward)

vₓ = 15 m/s (unchanged)

Speed = √(15² + 28²) = √(225 + 784) = √1009 = 31.8 m/s

Angle below horizontal: tanα = 28/15, so α = arctan(28/15) = 61.8°

The Equation of the Trajectory

Eliminating t from the parametric equations gives the Cartesian equation of the path:

From x = (ucosθ)t, we get t = x/(ucosθ).

Substituting into y = (usinθ)t - ½gt²:

y = xtanθ - gx² / (2u²cos²θ)

This is a parabola (y is a quadratic function of x). It is useful for finding the height at a given horizontal distance, or the horizontal distance at a given height.

Worked Example: A ball is launched at 25 m/s at 60° above the horizontal. Find its height when it has travelled 20 m horizontally.

y = 20tan60° - (9.8)(20²) / (2(25²)cos²60°)

= 20√3 - (9.8)(400) / (2(625)(0.25))

= 34.64 - 3920/312.5

= 34.64 - 12.54

= 22.1 m

Two Projectiles and Interception

When two projectiles are launched at different times or from different positions, set up position equations for each and solve simultaneously to find when (and if) they are at the same point.

Exam Tips

• Always separate the problem into horizontal and vertical components. Never mix them.
• The horizontal velocity is constant — there is no horizontal acceleration.
• If the projectile lands at a different height from the launch point, you cannot use the symmetric formula T = 2usinθ/g. Solve the vertical equation for the actual landing height.
• At any instant, the speed is √(vₓ² + v_y²) and the direction is arctan(v_y/vₓ) to the horizontal.
• In trajectory questions, check whether the question gives the launch angle or whether you need to find it.
• State your sign convention (e.g., upward positive) clearly and stick to it throughout.