Question 18
An object is projected at an acute angle of $\theta$ below the horizontal, with an initial speed of 30 m s−1 from a position 90 m above ground level.
The object hits the ground 90 m horizontally from its projection point.
Use vector calculus to determine $\theta$ in its simplest form.
Assume that the magnitude of mean acceleration due to gravity on Earth is 10 m s−2 and that there is no air resistance.
The object drops 90 m while it travels 90 m across
Watch where the object lands. You are told both distances, and you have to find the angle that makes them work.
The drawing is to scale, so you can see that the drop and the distance really are equal.
Change the angle until the object lands on the 90 m mark
Move the slider to change $\theta$. You will find two angles that hit the mark, but only one of them points below the horizontal.
Work through the solution one mark at a time
Try each step yourself before you reveal it. This follows the QCAA's first method, with downwards as positive and the origin at the release point.
You start from the acceleration and integrate twice. Take $\hat{\mathbf{i}}$ as horizontal and $\hat{\mathbf{j}}$ as vertical, with downwards as the positive direction, so gravity is positive and your launch velocity has a positive $\hat{\mathbf{j}}$ component.
It is worth being clear about what this does to $\theta$. You have taken the usual reference and flipped it upside down. Normally you measure $\theta$ upwards from the horizontal, so an angle above the horizontal is positive. With $\hat{\mathbf{j}}$ pointing down, you measure $\theta$ downwards instead, so $\theta=0$ is along the horizontal and $\theta=\frac{\pi}{2}$ is straight down. That is why an acute $\theta$ means the object is heading down, and why $30\sin(\theta)$ comes out positive.
$$\mathbf{a}(t)=10\,\hat{\mathbf{j}}\qquad \mathbf{v}(t)=\int \mathbf{a}(t)\,dt=10t\,\hat{\mathbf{j}}+\mathbf{c}$$
The launch velocity gives you the constant.
$$\mathbf{v}(0)=30\cos(\theta)\,\hat{\mathbf{i}}+30\sin(\theta)\,\hat{\mathbf{j}}$$
$$\mathbf{v}(t)=30\cos(\theta)\,\hat{\mathbf{i}}+\big(30\sin(\theta)+10t\big)\,\hat{\mathbf{j}}$$
Integrate again. If you put the origin at the release point, so that $\mathbf{r}(0)=0\,\hat{\mathbf{i}}+0\,\hat{\mathbf{j}}$, the second constant is zero.
$$\mathbf{r}(t)=30\cos(\theta)\,t\,\hat{\mathbf{i}}+\big(30\sin(\theta)\,t+5t^{2}\big)\,\hat{\mathbf{j}}$$
Upwards as positive is fine too, and so is an origin on the ground below the cliff. You just need to stay consistent once you have chosen.
Marker: correctly determines the position vector of the object showing evidence of the use of vector calculus
When the object hits the ground it is 90 m across and 90 m down from your origin. You set each component of $\mathbf{r}(t)$ equal to 90.
$$30\cos(\theta)\,t=90\quad\ldots(1)$$
$$30\sin(\theta)\,t+5t^{2}=90\quad\ldots(2)$$
You can simplify these to $\cos(\theta)\,t=3$ and $6\sin(\theta)\,t+t^{2}=18$ if you prefer. This mark can be implied by your later working, and you keep follow-through marks if an earlier slip carried into here.
Marker: determines two simultaneous equations in terms of $\theta$ and $t$
You need to get rid of $t$. Equation (1) gives you $t$ straight away, so substitute it into equation (2).
$$t=\frac{3}{\cos(\theta)}$$
$$30\sin(\theta)\left(\frac{3}{\cos(\theta)}\right)+5\left(\frac{3}{\cos(\theta)}\right)^{2}=90$$
Marker: uses simultaneous equations to determine equation in terms of $\theta$
This is where you have to spot the identity. The first term becomes $90\tan(\theta)$ and the second becomes $45\sec^{2}(\theta)$. Then $\sec^{2}(\theta)=1+\tan^{2}(\theta)$ gives you a quadratic in $\tan(\theta)$.
$$90\tan(\theta)+45\sec^{2}(\theta)=90$$
$$2\tan(\theta)+1+\tan^{2}(\theta)=2$$
$$\tan^{2}(\theta)+2\tan(\theta)-1=0$$
There is a second route you might like more. If you multiply through by $\cos^{2}(\theta)$ and use the double angle identities, you get $\tan(2\theta)=1$ instead. On that route, this mark is for your equation in $2\theta$.
Marker: determines equation in terms of $\tan(\theta)$
Use the quadratic formula with $\tan(\theta)$ as the unknown. You should get two values.
$$\tan(\theta)=\frac{-2\pm\sqrt{4-4\times1\times(-1)}}{2}=\frac{-2\pm2\sqrt{2}}{2}=-1\pm\sqrt{2}$$
On the double angle route, this mark is for a value of $2\theta$ such as $\frac{\pi}{4}$. Either way, you can have it implied by your next line.
Marker: determines both possible values of $\tan(\theta)$
Now you decide which value makes sense. The angle is acute, so $\tan(\theta)$ must be positive and you reject $-1-\sqrt{2}$. Write that reason down, because the reason is what earns this mark.
$\theta=\tan^{-1}\!\left(\sqrt{2}-1\right)=\dfrac{\pi}{8}$
You can give $\tan^{-1}(\sqrt{2}-1)$, $\frac{\pi}{8}$ or $22.5^{\circ}$. You should avoid a rounded radian value such as 0.393, because that is not in simplest form. If you try the rejected root in the interactive above, you will see it is a real path. It is a lob at 67.5° above the horizontal, and the question rules that out.
Marker: determines $\theta$ by evaluating the reasonableness of the solution
Each phrase of the question gave you something
Most projectile questions give you the angle and ask where the object lands. This one runs backwards. You are given the landing point and asked for $\theta$, so $\theta$ sits inside both components at once and you have to eliminate $t$ before you can find it.
If I had to guess where you might lose marks, it would be step 4. You have to notice that dividing by $\cos(\theta)$ leaves you with $\tan(\theta)$ and $\sec^{2}(\theta)$, and then reach for the Pythagorean identity. The QCAA like to finish with a reasonableness mark, so you also need to say in words why you rejected $-1-\sqrt{2}$.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and marking-guide steps are from the 2025 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2025, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.