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.

Resolving the initial velocity

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

DMATH_0

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:

DMATH_3
DMATH_4

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.

Worked example with numbers

A ball is kicked from ground level at $v_0 = 20$ m/s at $\theta = 35°$ above horizontal. Find the maximum height, time of flight, and range.

Resolve: $v_{0x} = 20 \cos 35° = 16.38$ m/s, $v_{0y} = 20 \sin 35° = 11.47$ m/s.

Maximum height: $h_{\max} = \frac{v_{0y}^2}{2g} = \frac{11.47^2}{19.6} = 6.71$ m.

Time of flight: $t = \frac{2 v_{0y}}{g} = \frac{22.94}{9.8} = 2.34$ s.

Range: $R = v_{0x} t = 16.38 \times 2.34 = 38.3$ m.

Try it: Projectile motion calculator - enter launch speed, angle and height and get the range, max height and trajectory.

Common traps

Mixing up horizontal and vertical equations. Horizontal velocity never changes (in HSC, where we ignore air resistance). Vertical velocity changes by $-9.8 \text{ m/s}$ every second. Set up two separate columns of working.

Forgetting the sign of $g$. If you take up as positive, $g$ enters the SUVAT equations as $-9.8 \text{ m/s}^2$. If you take down as positive, $g$ is $+9.8 \text{ m/s}^2$. Pick a convention and stick to it for the whole question.

Using the speed instead of a component. $v_0 = 25$ m/s at $40°$ does not mean the horizontal velocity is $25$ m/s. You must resolve into components first.

Treating horizontally thrown objects as having $v_{0y} = v_0$. If a stone is thrown horizontally off a cliff, $v_{0y} = 0$. The full speed is in the horizontal direction.

Forgetting units. Markers deduct for missing units (m, s, m/s) even when the number is correct.

In one sentence

Projectile motion is solved by splitting the initial velocity into horizontal and vertical components and applying constant-velocity equations horizontally and SUVAT equations vertically, linked only by the shared time of flight.