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Year 12 Specialist (Unit 3 & 4) Vector calculus

Projectile motion

20 practice questions 0 video lessons Theory + worked examples

Study projectile motion as vector calculus for Year 12 Specialist Mathematics in Queensland (QCAA). With gravity the only force, the acceleration is the constant vector −g j; integrating it once gives the velocity and again gives the position of the object at any time, taking g as \(9.8\ \text{m/s}^2\).

You will learn to resolve a launch velocity into components, integrate the acceleration to build \(\mathbf{v}(t)\) and \(\mathbf{r}(t)\), and find the time of flight, range, maximum height and the parabolic path — core modelling skills in Unit 3 vector calculus.

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Theory

Projectile motion applies vector calculus in Year 12 Specialist Mathematics (QCAA, Queensland). With gravity the only force, the acceleration is the constant vector \(\mathbf{a}=-g\mathbf{j}\); integrating it gives the velocity \(\mathbf{v}(t)\) and integrating again gives the position \(\mathbf{r}(t)\). From these you find the time of flight, range, maximum height and the parabolic path. Throughout, take \(g=9.8\ \text{m/s}^2\).

A projectile is an object moving under gravity alone, with air resistance ignored. Taking \(\mathbf{i}\) horizontal and \(\mathbf{j}\) vertically up, the only acceleration is gravity, so \(\mathbf{a}=-g\mathbf{j}\) with \(g=9.8\ \text{m/s}^2\). There is no horizontal acceleration, so the horizontal velocity stays constant.

The motion is built by vector calculus: velocity is the integral of acceleration, and position is the integral of velocity, with the constants of integration fixed by the initial velocity \(\mathbf{v}(0)\) and the initial position \(\mathbf{r}(0)\). Differentiating reverses this: \(\mathbf{v}=\dot{\mathbf{r}}\) and \(\mathbf{a}=\dot{\mathbf{v}}\), each done component by component.

A launch of speed \(u\) at angle \(\theta\) above the horizontal is first resolved into components: the horizontal part is \(u\cos\theta\) and the vertical part is \(u\sin\theta\). These are the components of \(\mathbf{v}(0)\).

Key quantities follow from the vertical motion: the maximum height occurs when the vertical velocity is zero, the time of flight is when the object returns to its launch height, and the range is the horizontal distance travelled in that time. Eliminating \(t\) between \(x(t)\) and \(y(t)\) gives the Cartesian equation of the path, which is always a parabola.

Projectile trajectory A projectile launched from the origin follows a parabolic path. The launch angle theta is marked at the origin, the apex is the maximum height H at the middle of the flight, the horizontal distance to landing is the range R, and the acceleration arrow points straight down (a equals minus g j). x (m) y (m) H R θ a = -g j
Parabolic path: the apex is the maximum height \(H\), \(R\) is the range, and the acceleration \(\mathbf{a}=-g\mathbf{j}\) points straight down throughout.
Resolving the launch velocity The launch velocity u at angle theta above the horizontal is the hypotenuse of a right-angled triangle. Its horizontal component is u cosine theta and its vertical component is u sine theta. θ u u cosθ u sinθ
Resolving the launch velocity \(u\) at angle \(\theta\) into a horizontal part \(u\cos\theta\) and a vertical part \(u\sin\theta\).

Start from the acceleration and integrate. With initial velocity \(\mathbf{v}(0)=u\cos\theta\,\mathbf{i}+u\sin\theta\,\mathbf{j}\) and launch from the origin:

\[ \mathbf{a}=-g\mathbf{j},\qquad \mathbf{v}(t)=u\cos\theta\,\mathbf{i}+(u\sin\theta-gt)\mathbf{j} \]
v(t)=ucosθi+(usinθ-gt)j

Integrating the velocity gives the position vector:

\[ \mathbf{r}(t)=u\cos\theta\,t\,\mathbf{i}+\left(u\sin\theta\,t-\tfrac{1}{2}gt^{2}\right)\mathbf{j} \]
r(t)=ucosθti+(usinθt-12gt2)j

The standard results (level ground) come from the vertical component:

\[ t_{\text{apex}}=\dfrac{u\sin\theta}{g},\qquad H=\dfrac{(u\sin\theta)^{2}}{2g},\qquad T=\dfrac{2u\sin\theta}{g},\qquad R=\dfrac{u^{2}\sin 2\theta}{g} \]
H=(usinθ)²2g

Eliminating \(t\) from \(x=u\cos\theta\,t\) gives the parabolic path:

\[ y=x\tan\theta-\dfrac{g\,x^{2}}{2u^{2}\cos^{2}\theta} \]
Split the motion. Horizontal: constant velocity, \(x=u\cos\theta\,t\). Vertical: constant acceleration \(-g\), \(y=u\sin\theta\,t-\tfrac{1}{2}gt^{2}\). The two are linked only through the shared time \(t\).

Solving a projectile problem

  1. Resolve the launch velocity into components: \(u\cos\theta\) horizontal and \(u\sin\theta\) vertical, giving \(\mathbf{v}(0)\).
  2. Integrate \(\mathbf{a}=-g\mathbf{j}\) to get \(\mathbf{v}(t)\) (apply \(\mathbf{v}(0)\)), then integrate again for \(\mathbf{r}(t)\) (apply \(\mathbf{r}(0)\)).
  3. Use the vertical part: set \(v_y=0\) for the maximum height, or \(y=0\) for the landing time (time of flight).
  4. Substitute back the time to find the range or height, or eliminate \(t\) between \(x\) and \(y\) for the Cartesian equation of the path.
Example 1 — Resolve the launch velocity
A ball is kicked from level ground at \(20\ \text{m/s}\) at \(30^\circ\) above the horizontal. Write its initial velocity vector (\(\mathbf{i}\) horizontal, \(\mathbf{j}\) up).
Solution

Split the speed into horizontal and vertical parts:

\(v_x\)\(=\)\(20\cos 30^\circ\)
\(=\)\(10\sqrt{3}\)
\(v_y\)\(=\)\(20\sin 30^\circ\)
\(=\)\(10\)
\(\mathbf{v}(0)\)\(=\)\(10\sqrt{3}\,\mathbf{i}+10\,\mathbf{j}\)

\(\mathbf{v}(0)=10\sqrt{3}\,\mathbf{i}+10\,\mathbf{j}\ \text{m/s}\).

Example 2 — Integrate to v(t) and r(t)
A ball leaves the origin with \(\mathbf{v}(0)=14\mathbf{i}+19.6\mathbf{j}\ \text{m/s}\) and \(\mathbf{a}=-9.8\mathbf{j}\ \text{m/s}^2\). Find \(\mathbf{v}(t)\), \(\mathbf{r}(t)\), and the time to maximum height.
Solution

Integrate the acceleration; the constant is \(\mathbf{v}(0)\):

\(\mathbf{v}\)\(=\)\(\textstyle\int(-9.8\mathbf{j})\,dt\)
\(=\)\(-9.8t\,\mathbf{j}+\mathbf{c}\)
\(\mathbf{v}(t)\)\(=\)\(14\mathbf{i}+(19.6-9.8t)\mathbf{j}\)

Integrate again; the constant is \(\mathbf{r}(0)=\mathbf{0}\):

\(\mathbf{r}(t)\)\(=\)\(14t\,\mathbf{i}+(19.6t-4.9t^{2})\mathbf{j}\)

Maximum height is where the vertical velocity is zero:

\(19.6-9.8t\)\(=\)\(0\)
\(t\)\(=\)\(2\)

\(\mathbf{v}(t)=14\mathbf{i}+(19.6-9.8t)\mathbf{j}\), \(\mathbf{r}(t)=14t\,\mathbf{i}+(19.6t-4.9t^{2})\mathbf{j}\); apex at \(t=2\ \text{s}\).

Example 3 — Time of flight and range
For the same ball (\(\mathbf{r}(t)=14t\,\mathbf{i}+(19.6t-4.9t^{2})\mathbf{j}\)), find the time of flight and the range on level ground.
Solution

The ball lands when its height returns to zero:

\(19.6t-4.9t^{2}\)\(=\)\(0\)
\(4.9t(4-t)\)\(=\)\(0\)
\(t\)\(=\)\(0\ \text{or}\ 4\)

The range is the horizontal distance at \(t=4\):

\(x\)\(=\)\(14\times 4\)
\(=\)\(56\)

Time of flight \(4\ \text{s}\); range \(56\ \text{m}\).

Projectile trajectory A projectile launched from the origin follows a parabolic path. The launch angle theta is marked at the origin, the apex is the maximum height H at the middle of the flight, the horizontal distance to landing is the range R, and the acceleration arrow points straight down (a equals minus g j). x (m) y (m) H R θ a = -g j
Example 4 — Cartesian path and clearing a wall
A ball leaves the origin with \(\mathbf{r}(t)=14t\,\mathbf{i}+(19.6t-4.9t^{2})\mathbf{j}\). Find the Cartesian equation of its path, and check whether it clears a \(5\ \text{m}\) wall \(50\ \text{m}\) away.
Solution

Write \(x\) in terms of \(t\), then eliminate \(t\):

\(x\)\(=\)\(14t \;\Rightarrow\; t=\dfrac{x}{14}\)
\(y\)\(=\)\(19.6\!\left(\dfrac{x}{14}\right)-4.9\!\left(\dfrac{x}{14}\right)^{2}\)
\(=\)\(1.4x-\dfrac{x^{2}}{40}\)

Find the height of the path at the wall \((x=50)\):

\(y(50)\)\(=\)\(1.4(50)-\dfrac{50^{2}}{40}\)
\(=\)\(70-62.5\)
\(=\)\(7.5\)

Compare with the wall height:

\(7.5\)\(>\)\(5\)
\(7.5-5\)\(=\)\(2.5\)

Path \(y=1.4x-\dfrac{x^{2}}{40}\); it clears the wall by \(2.5\ \text{m}\).

Common pitfalls

Forgetting to resolve the launch velocity. The launch speed \(u\) is not the horizontal or vertical velocity by itself — split it first into \(u\cos\theta\) and \(u\sin\theta\) before using the equations of motion.
Putting acceleration on the horizontal. Gravity acts only downward, so \(\mathbf{a}=-g\mathbf{j}\). The horizontal velocity has no acceleration and stays constant for the whole flight.
Mixing up time to the apex with time of flight. The apex is at \(t=\dfrac{u\sin\theta}{g}\); on level ground the total flight time is twice that. Do not use one where the other is needed.
Dropping the constant of integration. Each integration adds a constant that is fixed by the initial condition — \(\mathbf{v}(0)\) for velocity and \(\mathbf{r}(0)\) for position. Leaving it out loses the launch speed or the launch point.

Frequently asked questions

Why is the horizontal velocity constant in projectile motion?

Gravity acts only vertically, so \(\mathbf{a}=-g\mathbf{j}\) has no horizontal part. With no horizontal acceleration, the horizontal velocity \(u\cos\theta\) never changes.

What value of g is used, and how is it treated?

Take \(g=9.8\ \text{m/s}^2\) directed downward, so the acceleration vector is \(\mathbf{a}=-9.8\mathbf{j}\). State it each time you set up the problem.

How do you find the velocity and position from the acceleration?

Integrate \(\mathbf{a}=-g\mathbf{j}\) with respect to time to get \(\mathbf{v}(t)\), using \(\mathbf{v}(0)\) for the constant; integrate \(\mathbf{v}(t)\) again for \(\mathbf{r}(t)\), using \(\mathbf{r}(0)\).

How do you find the maximum height and the time of flight?

The maximum height is where the vertical velocity is zero, \(u\sin\theta-gt=0\); the time of flight (level ground) is when the height returns to zero, \(y=0\).

What shape is the path of a projectile?

A parabola. Eliminating \(t\) between \(x=u\cos\theta\,t\) and \(y=u\sin\theta\,t-\tfrac{1}{2}gt^2\) gives \(y=x\tan\theta-\dfrac{gx^2}{2u^2\cos^2\theta}\).

How do you find the speed and direction at a given time?

Substitute the time into \(\mathbf{v}(t)\); the speed is the magnitude \(|\mathbf{v}|=\sqrt{v_x^2+v_y^2}\) and the direction is \(\tan^{-1}\!\left(\dfrac{v_y}{v_x}\right)\) from the horizontal.

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