Question 17
An object is projected upwards from ground level with an initial velocity of 15 $\text{m s}^{-1}$ at an angle of $54^\circ$ to the horizontal.
The object just passes over a drone hovering in the air. An observer is positioned directly below the drone and at a horizontal distance of 20 m from where the object is projected.
The observer commented that:
- it took the object around 2 to 2.5 seconds after its projection to reach the drone
- the object was still moving in an upwards direction as it passed the drone.
Assuming that air resistance is negligible, use a vector calculus approach to evaluate the reasonableness of the observer’s comments.
The timing checks out, but the object is already on its way down
Watch the gauge on the right. It shows the vertical part of the velocity, and it crosses zero at the highest point, well before the object reaches the drone.
The object reaches the drone inside the observer’s 2 to 2.5 seconds. By then it has been falling for about a second, so the second comment does not hold.
Move through the flight and test each comment
Drag the time slider. The readouts give the position and velocity vectors at that moment, from the origin at the launch point, and the gauge shows the vertical part of the velocity. Each comment is a claim about one of these.
The $\hat{\jmath}$ part of the velocity is positive, so the object is still rising. It hasn’t reached the drone yet.
This is the highest point, near enough. The $\hat{\jmath}$ part of the velocity is zero at about $t=1.24$ s, so the readout here is almost 0. From now on the object moves down.
The $\hat{\jmath}$ part of the velocity is now negative, so the object is falling. It is still short of the drone at 20 m across.
The $\hat{\imath}$ part of the position is about 20, so the object is over the drone, at about $t=2.27$ s. That fits the first comment. The $\hat{\jmath}$ part of the velocity is about $-10.1$, so it is moving down and the second comment fails.
The object is past the drone and still falling. It lands at about $t=2.48$ s, about 21.8 m from the launch point.
The blue stretch is the observer’s window. It starts at 2 seconds, when the object is 17.6 m across. At 2.5 seconds it would be 22.0 m across, so the 20 m mark is crossed inside the window. The object actually lands at about 2.48 seconds, just before the window ends.
Work through the solution one mark at a time
Try each step yourself before you reveal it. These steps follow Method 1 of the marking guide. The other two methods are described after the last step.
Set up the model before you integrate. Let $\hat{\imath}$ and $\hat{\jmath}$ be the horizontal and vertical unit vectors, with $\hat{\jmath}$ pointing up, and let $t$ be the time in seconds after the projection. With no air resistance, gravity is the only acceleration.
$$\begin{aligned}\boldsymbol{a}(t)&=-9.8\hat{\jmath}\\\boldsymbol{v}(t)&=\int\boldsymbol{a}(t)\,dt\\&=-9.8t\hat{\jmath}+\boldsymbol{c}\end{aligned}$$
The constant is the launch velocity. Split 15 $\text{m s}^{-1}$ at $54^\circ$ into components.
$$\begin{aligned}\boldsymbol{v}(0)&=15\cos(54^\circ)\hat{\imath}\\&\quad+15\sin(54^\circ)\hat{\jmath}\\\boldsymbol{v}(t)&=15\cos(54^\circ)\hat{\imath}\\&\quad+\left(15\sin(54^\circ)-9.8t\right)\hat{\jmath}\end{aligned}$$
Check your calculator is in degree mode. An equivalent function is accepted, such as $8.8\hat{\imath}+(12.1-9.8t)\hat{\jmath}$, and so is velocity written in components, $v_x=15\cos(54^\circ)$ and $v_y=15\sin(54^\circ)-gt$.
Marker: correctly determines the velocity function of the object using vector calculus
Integrate again for the position vector.
$$\begin{aligned}&\boldsymbol{r}(t)=\int\boldsymbol{v}(t)\,dt\\&\quad=15\cos(54^\circ)t\hat{\imath}\\&\qquad+\left(15\sin(54^\circ)t-4.9t^2\right)\hat{\jmath}\\&\qquad+\boldsymbol{c}\end{aligned}$$
Put the origin at the release point, so $\boldsymbol{r}(0)=0\hat{\imath}+0\hat{\jmath}$ and the constant is zero.
$$\begin{aligned}&\boldsymbol{r}(t)=15\cos(54^\circ)t\hat{\imath}\\&\quad+\left(15\sin(54^\circ)t-4.9t^2\right)\hat{\jmath}\end{aligned}$$
Follow-through applies if your velocity had an error. An equivalent function is accepted, such as $8.8t\hat{\imath}+\left(12.1t-4.9t^2\right)\hat{\jmath}$.
Marker: determines displacement function of the object
The observer stands directly below the drone, 20 m from the launch point. So the object is over the drone when the $\hat{\imath}$ component of $\boldsymbol{r}(t)$ equals 20.
$$\begin{aligned}&15\cos(54^\circ)t=20\\&t=\frac{20}{15\cos(54^\circ)}\approx2.27\text{ s}\end{aligned}$$
An equivalent value is accepted, such as 2.3.
Marker: determines time when the object just passes drone
The object rises until it reaches its maximum height, and falls after that. Your calculator can find the maximum of the graph of $15\sin(54^\circ)t-4.9t^2$, or you can set the $\hat{\jmath}$ component of $\boldsymbol{v}(t)$ to zero.
$$\begin{aligned}&15\sin(54^\circ)-9.8t=0\\&t=\frac{15\sin(54^\circ)}{9.8}\approx1.24\text{ s}\end{aligned}$$
Another way to settle the direction also earns this mark. You can show that the vertical velocity at the drone is negative, for example $v_y(2.27)\approx-10.1$ or $v_y(2.27)<0$, or show it on a relevant graph.
Marker: determines time when the object reaches maximum height
Give each comment its own verdict and point to the number that decides it. The object passes the drone at about 2.27 s, which is inside 2 to 2.5 seconds. It reached its highest point at about 1.24 s, before it got to the drone, so it was already moving down when it passed.
An equivalent evaluation is accepted when it is based on your earlier working. The second mark is for how you communicate. The guide looks for technical vocabulary, a clear flow, and conventions such as vector notation and a stated origin.
The marking guide gives two other routes that earn the same marks. Method 2 works from the velocity alone. The horizontal distance is 17.64 m at 2 seconds and 22.05 m at 2.5 seconds, so the 20 m mark falls inside the window, and $v_y(2)=-7.46\text{ m s}^{-1}$ shows the object is already moving down at 2 seconds. Method 3 uses $\boldsymbol{r}(t)$ at both times, with $x=17.64$ and $y=4.67$ at 2 seconds and $x=22.05$ and $y=-0.29$ at 2.5 seconds. Both routes end with the same two verdicts.
Marker: uses mathematical justification to evaluate the reasonableness of both comments based on prior mathematical reasoning [1 mark]; shows logical organisation, communicating key steps [1 mark]
Each phrase of the question gave you something
Nothing is broken into parts. You build the vector model yourself, choosing the origin and the unit vectors, and then you have to decide which quantity tests each comment. The time at the drone checks the first one, and the time of maximum height or the sign of the vertical velocity checks the second. The question also asks you to judge claims, so your answer has to end in verdicts. It stays complex familiar because “use a vector calculus approach” tells you exactly how to model the flight, and every calculation along the way is one you have practised.
If I had to guess where you might lose marks, it would be the evaluation and the communication mark. The subject report used this question as an example of effective practice. The responses it highlights “made suitable assumptions regarding the position of the origin and the direction of the unit vectors used to model the motion of the object” and “used a vector calculus approach as required in the problem”. So state those assumptions before you integrate, and give each comment its own verdict backed by a number from your working.
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 2023 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2023, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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