Solve problems involving projectile motion by resolving the motion into independent horizontal and vertical components, assuming constant downward acceleration due to gravity and negligible air resistance

What this dot point is asking

QCAA wants you to model the motion of a projectile (an object moving only under gravity) by resolving its velocity into independent horizontal and vertical components, then applying the constant-acceleration equations to each axis. The two stated assumptions are constant downward acceleration $g = 9.8 \text{ m/s}^2$ and negligible air resistance. The dot point underpins every projectile calculation in IA1 and EA Paper 2, and is the classic IA2 design (range vs angle).

The answer

A projectile is any object in free flight subject only to gravity. The horizontal and vertical motions 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:

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Horizontal motion

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

Vertical motion

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

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Key features of the trajectory

The path is a parabola. At maximum height $v_y = 0$, so the rise above the launch point is:

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

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

If the launch and landing heights differ (release height above the ground, or landing on a raised platform), do not use the level-ground range formula. Set the displacement $\Delta y$ at landing explicitly and solve the vertical equation for $t$, then use $x = v_{0x} t$.

Worked example

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: $h_{\max} = v_{0y}^2 / (2g) = 11.47^2 / 19.6 = 6.71$ m.

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

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

Try it: Projectile motion calculator. Enter launch speed, angle and release height to get the range, max height and trajectory.

How this appears in IA1 and IA2

IA1 data test. Expect either a launch-angle table to interpret (range vs angle data, often missing one value), or a single trajectory with annotated heights and asked to extract launch speed and landing time. Marker traps focus on candidates who plug the resultant speed in place of a component or forget to use $r = R_E + h$ on the vertical axis when a height above the launch is involved.

IA2 student experiment. The standard projectile IA2 measures range as a function of launch angle for a fixed launch speed (small projectile launcher or a ball-bearing on a ramp). A strong report linearises by plotting $R$ against $\sin(2\theta)$ (slope $= v_0^2 / g$), reports a $v_0$ that agrees with a direct measurement, and discusses air-resistance and release-height systematic effects in the evaluation. The IA2 criteria reward the design justification, the data, and the evaluation in that order.

Common traps

Mixing up horizontal and vertical equations. Horizontal velocity never changes (with air resistance neglected). Vertical velocity changes by $-9.8$ m/s every second. Set up two separate columns of working and never let a $\cos$ leak into a vertical equation.

Forgetting the sign of $g$. If you take up as positive, $g$ enters the 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.

Using the level-ground range formula on uneven terrain. $R = v_0^2 \sin(2\theta) / g$ only works when launch height equals landing height. Otherwise solve the vertical equation for $t$ first, then compute $x = v_{0x} t$.

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 resolving the launch velocity into independent horizontal and vertical components, applying constant-velocity equations horizontally and constant-acceleration equations vertically (with $g$ down), and linking the two axes through a shared time of flight.