Analyse the motion of projectiles by resolving the motion into horizontal and vertical components, making the following assumptions: a constant vertical acceleration due to gravity, zero air resistance

What this dot point is asking

NESA wants you to model the motion of a projectile (an object moving only under gravity) by splitting its velocity into independent horizontal and vertical components, then applying the equations of motion to each axis. The two key assumptions are constant downward acceleration $g = 9.8 \text{ m/s}^2$ and no air resistance. This dot point underpins every calculation question in projectile motion and appears in some form in nearly every Module 5 exam.

The answer

A projectile is any object in flight that is subject only to gravity. The trick is that horizontal and vertical motion are independent, linked only by the shared time of flight. The diagram shows the trajectory with the labelled vectors and equations you need.

Resolving the initial velocity

If a projectile is launched with speed $v_0$ at angle $\theta$ above the horizontal:

Horizontal motion

No horizontal force acts (air resistance is ignored), so horizontal velocity is constant.

Vertical motion

The only acceleration is gravity, $a_y = -g$ (taking up as positive). Use SUVAT:

Key features of the trajectory

The path is a parabola. At maximum height $v_y = 0$, so $h_{\max} = \frac{v_{0y}^2}{2g}$.

For a projectile launched from and landing at the same height, the time of flight is $t = \frac{2 v_{0y}}{g}$ and the range is:

Range is maximised at $\theta = 45°$ (for level ground), and complementary angles (for example, $30°$ and $60°$) give the same range.

Examples in context

Example 1. Cricket throw from the SCG outfield. A fielder at deep mid-wicket flicks the ball at $v_0 = 28 \text{ m/s}$ at $\theta = 32^{\circ}$ above horizontal toward the keeper. Components are $v_{0x} = 28 \cos 32^{\circ} = 23.74 \text{ m/s}$ and $v_{0y} = 28 \sin 32^{\circ} = 14.84 \text{ m/s}$. Time of flight back to the same height is $t = 2 v_{0y} / g = 2 \times 14.84 / 9.8 = 3.03 \text{ s}$, and the horizontal carry is $R = v_{0x} t = 23.74 \times 3.03 = 71.9 \text{ m}$. The keeper, 65 m away, sees the ball arrive on the bounce. Air drag makes the real carry $\sim 8\%$ shorter, but the components method gives the right ballpark.

Example 2. Cliff jumper at Coffs Harbour Mutton Bird Island. A diver runs off a $14.0 \text{ m}$ cliff horizontally at $v_x = 4.5 \text{ m/s}$. Vertical motion starts from rest, so $14.0 = \tfrac{1}{2} g t^2$ gives $t = \sqrt{2 \times 14.0 / 9.8} = 1.69 \text{ s}$. Horizontal distance from the base is $x = v_x t = 4.5 \times 1.69 = 7.6 \text{ m}$, well clear of the rock ledge. Final vertical speed at impact is $v_y = g t = 9.8 \times 1.69 = 16.6 \text{ m/s}$, giving impact speed $|\vec{v}| = \sqrt{4.5^2 + 16.6^2} = 17.2 \text{ m/s}$ at $\arctan(16.6/4.5) = 74.8^{\circ}$ below horizontal.

Try this

Q1. Define the term "projectile" as used in HSC Physics and state the two simplifying assumptions made about its motion. [2 marks]

• Cue. Object in flight subject only to gravity; constant vertical acceleration $g$; no air resistance. One mark per assumption.

Q2. A stone is thrown from the Sydney Harbour Bridge deck $49.0 \text{ m}$ above the water at $v_0 = 18.0 \text{ m/s}$ angled $25^{\circ}$ below horizontal. Calculate the horizontal distance to the splashdown. [4 marks]

• Cue. Resolve $v_{0y} = -18 \sin 25^{\circ}$ (down positive convention), solve $49 = |v_{0y}| t + \tfrac{1}{2} g t^2$ for $t$, then $x = 18 \cos 25^{\circ} \cdot t$.

Q3. A motocross rider leaves a $5.0^{\circ}$ ramp at $20 \text{ m/s}$. (a) Find the time spent airborne if landing height equals launch height. (b) Find the maximum height above the ramp. (c) Explain why doubling the launch angle to $10^{\circ}$ more than doubles the range. [2+2+2 marks]

• Cue. (a) $t = 2 v_0 \sin\theta / g$. (b) $h = v_0^2 \sin^2\theta / (2g)$. (c) Range $\propto \sin(2\theta)$; $\sin 20^{\circ} / \sin 10^{\circ} \approx 1.97$, so range nearly doubles.