Projectile motion is the motion of an object (a projectile) launched into the air and moving under the sole influence of gravity. In VCE Specialist Mathematics, this is analyzed using vectors and calculus, typically assuming no air resistance and a constant gravitational acceleration.
To analyze projectile motion, we establish a 2D Cartesian coordinate system:
* The origin \((0,0)\) is usually the point of projection or ground level.
* The unit vector \(\mathbf{i}\) represents the horizontal direction.
* The unit vector \(\mathbf{j}\) represents the vertical direction (upwards is positive).
* Gravity (\(g\)) acts vertically downwards: \(\mathbf{a} = -g\mathbf{j}\), where \(g \approx 9.8 \text{ m/s}^2\).
If a projectile is launched from an initial height \(h\) with an initial speed \(u\) at an angle \(\theta\) to the horizontal:
* Initial Position: \(\mathbf{r}(0) = 0\mathbf{i} + h\mathbf{j}\)
* Initial Velocity: \(\mathbf{v}(0) = u\cos(\theta)\mathbf{i} + u\sin(\theta)\mathbf{j}\)
REMEMBER: Always resolve the initial velocity into components immediately: \(v_x = u\cos(\theta)\) and \(v_y = u\sin(\theta)\).
Using the constant acceleration \(\mathbf{a}(t) = -g\mathbf{j}\), we use integration to find velocity and position.
Integrating acceleration with respect to time \(t\):
\$\(\mathbf{v}(t) = \int \mathbf{a}(t) \, dt = \int -g\mathbf{j} \, dt = -gt\mathbf{j} + \mathbf{c}\)\$
Using the initial condition \(\mathbf{v}(0) = u\cos(\theta)\mathbf{i} + u\sin(\theta)\mathbf{j}\):
\$\(\mathbf{v}(t) = (u\cos(\theta))\mathbf{i} + (u\sin(\theta) - gt)\mathbf{j}\)\$
Integrating velocity with respect to time \(t\):
\$\(\mathbf{r}(t) = \int \mathbf{v}(t) \, dt = \int \left[ (u\cos(\theta))\mathbf{i} + (u\sin(\theta) - gt)\mathbf{j} \right] \, dt\)\$
\$\(\mathbf{r}(t) = (u\cos(\theta)t)\mathbf{i} + \left( u\sin(\theta)t - \frac{1}{2}gt^2 + h \right)\mathbf{j}\)\$
The components of the position vector give the parametric equations:
* \(x(t) = u\cos(\theta)t\)
* \(y(t) = u\sin(\theta)t - \frac{1}{2}gt^2 + h\)
EXAM TIP: If a question asks for the “position of the particle,” provide a vector \(\mathbf{r}(t)\). If it asks for the “coordinates,” provide \((x, y)\).
Analyzing the motion involves finding specific values for time (\(t\)), horizontal distance (\(x\)), and vertical height (\(y\)).
| Feature | Condition | Method to Solve |
|---|---|---|
| Maximum Height | Vertical velocity is zero: \(\dot{y}(t) = 0\) | Solve \(u\sin(\theta) - gt = 0\) for \(t\), then substitute into \(y(t)\). |
| Time of Flight | Projectile hits the ground: \(y(t) = 0\) | Solve the quadratic \(u\sin(\theta)t - \frac{1}{2}gt^2 + h = 0\) for \(t > 0\). |
| Horizontal Range | Value of \(x(t)\) at the end of flight | Find time of flight \(T\), then calculate \(x(T)\). |
| Impact Velocity | Velocity at time of flight \(T\) | Calculate \(\mathbf{v}(T) = \dot{x}(T)\mathbf{i} + \dot{y}(T)\mathbf{j}\). |
KEY TAKEAWAY: The horizontal component of velocity \(v_x = u\cos(\theta)\) remains constant throughout the entire flight because there is no horizontal acceleration.
The path of a projectile is a parabola. To find the Cartesian equation, we eliminate the parameter \(t\) from the parametric equations.
General Cartesian Equation:
\$\(y = x\tan(\theta) - \frac{gx^2}{2u^2}(1 + \tan^2(\theta)) + h\)\$
VCAA FOCUS: This equation is frequently used to find the required launch angle \(\theta\) to hit a specific target \((x, y)\). It results in a quadratic equation in terms of \(\tan(\theta)\).
The speed of the projectile at any time \(t\) is the magnitude of the velocity vector:
\$\(v = |\mathbf{v}(t)| = \sqrt{(\dot{x})^2 + (\dot{y})^2}\)\$
The angle of motion \(\phi\) (the direction of the velocity vector) relative to the positive \(x\)-axis is:
\$\(\tan(\phi) = \frac{\dot{y}}{\dot{x}}\)\$
COMMON MISTAKE: Students often confuse “velocity” with “speed.” Velocity is a vector (e.g., \(3\mathbf{i} - 4\mathbf{j}\)), while speed is a scalar magnitude (e.g., \(\sqrt{3^2 + (-4)^2} = 5\)).
| Quantity | Formula |
|---|---|
| Time to Max Height | \(t = \frac{u\sin(\theta)}{g}\) |
| Maximum Height | \(H = \frac{u^2\sin^2(\theta)}{2g}\) |
| Total Time of Flight | \(T = \frac{2u\sin(\theta)}{g}\) |
| Horizontal Range | \(R = \frac{u^2\sin(2\theta)}{g}\) |
STUDY HINT: The range formula \(R = \frac{u^2\sin(2\theta)}{g}\) only applies if the projectile starts and ends at the same vertical height (\(h=0\)). Do not use it if the projectile is launched from a cliff or hits an elevated platform!
