Analyse projectile motion by treating the horizontal and vertical components independently, and determine range, maximum height and time of flight

What this dot point is asking

SEAB wants you to analyse the two-dimensional motion of a projectile by separating it into independent horizontal and vertical components, and to find quantities such as time of flight, maximum height and horizontal range. The key physical idea is that the only force (neglecting air resistance) is gravity, which acts vertically and leaves the horizontal motion unaffected.

The answer

The independence principle

A projectile experiences only its weight (air resistance neglected). Weight acts vertically downward, so:

• Horizontal motion has zero acceleration: the horizontal velocity is constant.
• Vertical motion has acceleration $g$ downward: it is uniformly accelerated motion.

These two motions are independent and share only one quantity: time. This is why the path is a parabola.

Resolving the launch velocity

For a launch speed $u$ at angle $\theta$ above the horizontal:

The horizontal component stays constant; the vertical component changes under gravity exactly as in vertical kinematics.

Time of flight

For a projectile launched and landing at the same height, the vertical displacement returns to zero:

For motion that lands at a different height, solve the full quadratic $s_y = u_y t - \tfrac{1}{2}g t^2$ for $t$.

Maximum height

At the highest point the vertical velocity is zero:

Horizontal range

The range is the constant horizontal velocity multiplied by the time of flight:

(using $2\sin\theta\cos\theta = \sin 2\theta$). The range is greatest at $\theta = 45^\circ$, and complementary angles (such as $30^\circ$ and $60^\circ$) give the same range.

The effect of air resistance

Real projectiles meet air resistance, which acts opposite to velocity. The path becomes asymmetric: the range and maximum height are reduced, the descent is steeper than the ascent, and the projectile lands at a steeper angle and lower speed than it launched. The trajectory is no longer a true parabola.

Examples in context

Example 1. Why complementary angles share a range. A ball launched at $30^\circ$ and one at $60^\circ$ with the same speed land at the same horizontal distance, because $\sin 2\theta$ takes the same value at $60^\circ$ and $120^\circ$. The $60^\circ$ launch goes higher and stays airborne longer but has a smaller horizontal velocity, so the products match.

Example 2. A long jump. A long jumper leaving the board at roughly $45^\circ$ maximises range in the idealised model, but in practice athletes launch nearer $20^\circ$ because they cannot generate the same speed at steep angles. This shows that the $45^\circ$ result assumes a fixed launch speed, an assumption that breaks down for the human body.

Try this

Q1. State why the horizontal velocity of a projectile (with air resistance neglected) is constant. [2 marks]

• Cue. The only force is weight, which acts vertically; there is no horizontal force, so horizontal acceleration and hence horizontal velocity are unchanged.

Q2. A projectile is launched at $18\ \text{m s}^{-1}$ at $50^\circ$ above the horizontal from level ground. Find its time of flight. [2 marks]

• Cue. $u_y = 18\sin 50^\circ = 13.8\ \text{m s}^{-1}$; $t = \dfrac{2 \times 13.8}{9.81} = 2.81\ \text{s}$.

Q3. Describe how the trajectory of a real projectile differs from the ideal parabolic path, and explain why. [3 marks]

• Cue. Air resistance opposes motion, reducing range and height, steepening the descent, and breaking the symmetry; the landing speed and angle differ from launch.