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

What this dot point is asking

VCAA wants you to model 2D projectile motion as two independent 1D problems linked by a shared time, applying the constant-acceleration equations to each axis with the stated assumptions of $g$ down and no air resistance.

Independence of axes

A projectile in free flight has only gravity acting. Horizontal acceleration is zero; vertical acceleration is $-g$ (taking up as positive). The horizontal and vertical motions are independent, connected only by the shared time of flight.

Resolving the initial velocity

For a launch at angle $\theta$ above the horizontal with speed $v_0$:

Horizontal motion (constant velocity)

Vertical motion (constant acceleration)

Key formulas

Max height (above launch): $h = v_{0y}^2 / (2g)$.

Time of flight (level ground): $T = 2 v_{0y} / g$.

Range (level ground): $R = v_0^2 \sin(2\theta) / g$. Maximum at $\theta = 45°$; complementary angles give the same range.

Off-level landing

If launch and landing heights differ (cliff drop or stepped target), do not use the level-ground range formula. Set the vertical displacement at landing explicitly and solve the vertical equation for $t$, then compute $x = v_{0x} t$.

Worked example (cliff drop)

A stone is thrown horizontally at $15$ m s$^{-1}$ from a $30$ m cliff. Find the time to reach the water and the horizontal distance.

$v_{0y} = 0$. Vertical: $y = \tfrac{1}{2} g t^2$ gives $t = \sqrt{2 y / g} = \sqrt{60/9.8} = 2.47$ s.

Horizontal: $x = v_{0x} t = 15 \cdot 2.47 = 37.1$ m.

Common traps

Mixing horizontal and vertical equations

Keep two separate columns of working.

Using the speed instead of a component

A launch at $25$ m s$^{-1}$ at $40°$ does not have horizontal velocity $25$ m s$^{-1}$.

Forgetting that for a horizontally launched object $v_{0y} = 0$

Not $v_0$.

Applying the level-ground range formula on uneven terrain

Only valid when launch height = landing height.

Try it: Projectile motion calculator.

In one sentence

Projectile motion is solved by resolving the launch velocity into horizontal ($v_{0x} = v_0 \cos\theta$) and vertical ($v_{0y} = v_0 \sin\theta$) 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.

Examples in context

Example 1. AFL torpedo punt at Marvel Stadium. A Marvel Stadium AFL torpedo is kicked at $25$ m s$^{-1}$ at $45^\circ$ above the horizontal. Horizontal velocity $v_x = 25 \cos 45^\circ = 17.7$ m s$^{-1}$; vertical $v_y = 25 \sin 45^\circ = 17.7$ m s$^{-1}$. Time to peak is $t_{\rm up} = v_y/g = 1.80$ s; total flight time $t = 2 t_{\rm up} = 3.60$ s; range $R = v_x t = 63.6$ m, slightly longer than the centre-square diagonal. Maximum height is $h = v_y^2/(2g) = 16.0$ m, just below the closed-roof ceiling. Air resistance (ignored here) reduces real-world range by approximately $5$ to $10$%.

Example 2. Bass Strait helicopter fire-cargo drop. A Bass Strait gas-platform supply helicopter, hovering at $50$ m altitude with $20$ m s$^{-1}$ forward velocity, drops a survival pod. Treat the pod as a projectile with $v_x = 20$ m s$^{-1}$, $v_y = 0$ initially. Time to splash: $50 = \tfrac{1}{2}g t^2$ gives $t = \sqrt{100/9.8} = 3.19$ s. Horizontal travel from release point to splash: $x = v_x t = 63.9$ m. Vertical impact speed: $v_y = gt = 31.3$ m s$^{-1}$. Total impact speed $v = \sqrt{20^2 + 31.3^2} = 37.1$ m s$^{-1}$, useful for designing impact-resistant packaging.

Try this

Q1. Explain why horizontal and vertical components of projectile motion can be treated independently. [2 marks]

• Cue. No horizontal force (neglecting air); only vertical gravitational force. Components share only the time variable.

Q2. A ball is kicked at $30$ m s$^{-1}$ at $60^\circ$ above horizontal. Calculate (a) the time of flight, and (b) the horizontal range. [4 marks]

• Cue. $v_y = 26.0$ m s$^{-1}$; $t = 2v_y/g = 5.30$ s. $v_x = 15.0$ m s$^{-1}$; $R = 15.0 \times 5.30 = 79.5$ m.

Q3. Refer to the helicopter cargo drop. (a) Calculate the time to splash. (b) Determine the horizontal distance travelled. (c) Find the impact speed. [2+2+2 marks]

• Cue. (a) $3.19$ s. (b) $63.9$ m. (c) $37.1$ m s$^{-1}$.