Question 14
An object is projected vertically upwards from ground level. After the object has been in motion for $t$ seconds, its position vector through the air, in metres, is modelled by
$$\boldsymbol{r}(t)=5t(8-t)\,\hat{\boldsymbol{\jmath}}$$
Determine the velocity of the object through the air, $\boldsymbol{v}(t)$, in metres per second.
[2 marks] b)Determine the number of seconds until the object reaches its maximum height.
[2 marks] c)Determine the maximum height that the object reaches, in metres.
[2 marks]The object stops for an instant at the top
On the left, the object goes straight up and comes back down. The blue arrow is its velocity. On the right, the graph plots its height against time.
The motion runs at real speed: $4$ seconds up and $4$ seconds down. The graph is the $\hat{\boldsymbol{\jmath}}$ component of $\boldsymbol{r}(t)$.
Move through the flight and watch the velocity
The top graph is the height, $5t(8-t)$, and the bottom graph is the velocity, $40-10t$. The blue tangent on the top graph has gradient equal to the velocity. The orange point is the maximum height.
At $t=0$ the object is at ground level, $\boldsymbol{r}(0)=0\,\hat{\boldsymbol{\jmath}}$, and its velocity is $40\,\hat{\boldsymbol{\jmath}}\ \text{m s}^{-1}$. That is the launch velocity.
The velocity is a positive multiple of $\hat{\boldsymbol{\jmath}}$, so the object is still going up. The tangent slopes upwards, and it gets flatter as the object slows down.
This is the top. The velocity is $0\,\hat{\boldsymbol{\jmath}}$ and the tangent is flat, so $t=4$ and the height is $80$ m.
The velocity is a negative multiple of $\hat{\boldsymbol{\jmath}}$, so the object is falling. Compare $t=4-k$ with $t=4+k$: the heights match and the velocities have opposite signs.
Back at ground level, $\boldsymbol{r}(8)=0\,\hat{\boldsymbol{\jmath}}$. It lands with velocity $-40\,\hat{\boldsymbol{\jmath}}\ \text{m s}^{-1}$, the same speed it was launched with.
Work through the solution one mark at a time
Try each step yourself before you reveal it. The steps follow Method 1 of the QCAA marking guide.
Expand the brackets first, so you can differentiate one term at a time.
$$\begin{aligned}\boldsymbol{r}(t)&=5t(8-t)\,\hat{\boldsymbol{\jmath}}\\&=\left(40t-5t^2\right)\hat{\boldsymbol{\jmath}}\end{aligned}$$
The QCAA accept the expansion without the $\hat{\boldsymbol{\jmath}}$. In Method 2 of the marking guide, this mark goes to the product rule instead, and it can be implied by your later working.
$$\begin{aligned}\boldsymbol{v}(t)&=\big(5t(-1)\\&\qquad+5(8-t)\big)\,\hat{\boldsymbol{\jmath}}\end{aligned}$$
Marker: correctly expands the given expression (Method 2: correctly uses the product rule to determine $\boldsymbol{v}(t)$)
Velocity is the derivative of position. Differentiate the $\hat{\boldsymbol{\jmath}}$ component and keep the unit vector.
$$\boldsymbol{v}(t)=\frac{d}{dt}\left(40t-5t^2\right)\hat{\boldsymbol{\jmath}}$$
Write the $\hat{\boldsymbol{\jmath}}$. The marking guide says it does not accept $40-10t$ on its own, because the question asks for the velocity vector. If you slipped in Step 1, you can still earn this mark by follow-through.
Marker: determines $\boldsymbol{v}(t)$
At its maximum height the object stops rising for an instant, so its velocity is zero. Set $\boldsymbol{v}(t)=0\,\hat{\boldsymbol{\jmath}}$ and compare the $\hat{\boldsymbol{\jmath}}$ components.
$$40-10t=0$$
The QCAA also accept an equivalent equation, such as $10t=40$.
Marker: establishes equation in terms of $t$
Solve for $t$.
$$10t=40$$
Here is a quick check. The height $5t(8-t)$ is zero at $t=0$ and $t=8$, and a parabola peaks halfway between its zeros.
Marker: determines $t$
Substitute your answer to part b), $t=4$, into the expression for position.
$$\begin{aligned}&\text{Maximum height}\\&\quad=5\times4(8-4)\end{aligned}$$
The QCAA also accept an equivalent substitution into the expanded form, such as $40\times4-5(4)^2$. This mark may be implied by your next line.
Marker: substitutes result from 14b) into expression for position
Evaluate.
$$5\times4(8-4)=20\times4=80$$
The position at the top is $\boldsymbol{r}(4)=80\,\hat{\boldsymbol{\jmath}}$, and the height is the size of that vector. The question asks for a number of metres, so give $80$ m.
Marker: determines maximum height
Each phrase of the question gave you something
The position is a quadratic in one direction, so the velocity takes one derivative. Each part then feeds the next: the velocity gives the time of the maximum, and that time gives the height.
If I had to guess where you might lose marks, it would be part a), where $40-10t$ without the $\hat{\boldsymbol{\jmath}}$ loses the second mark. The other place is part c): substituting $t=4$ into the velocity instead of the position gives $0$, which can’t be a maximum height.
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 2021 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2021, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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