2020HSCShort answer

Q12 (b) (i)

3 marks

Question

A particle is projected from the origin with initial velocity $u\,\mathrm{m/s}$ at an angle $\theta$ to the horizontal. The particle lands at $x=R$ on the $x$ -axis. The acceleration vector is
$\mathbf a=\begin{pmatrix}0\\-g\end{pmatrix},$
where $g$ is the acceleration due to gravity. (Do NOT prove this.)

(i) Show that the position vector $\mathbf r(t)$ of the particle is given by
$\mathbf r(t)=\begin{pmatrix}ut\cos\theta\\ut\sin\theta-\frac12gt^2\end{pmatrix}.$

Luna-requested uncropped source page for question 12.b.i; requires human review and cropping — Reviewed visual · source page 8

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Solution

Criteria	Marks
Provides correct solution	3
Correctly obtains $\vec v(t)=\begin{pmatrix}u\cos\theta\\u\sin\theta-gt\end{pmatrix}$ , or equivalent merit<br>OR<br>Correctly obtains one component of $\vec r(t)$	2
Correctly obtains one component of $\vec v(t)$ , or equivalent merit	1

Sample answer:

\vec a=\begin{pmatrix}0\\-g\end{pmatrix}

\vec v=\begin{pmatrix}c_1\\-gt+c_2\end{pmatrix}

Now at $t=0$ ,

\vec v=\begin{pmatrix}u\cos\theta\\u\sin\theta\end{pmatrix}

\therefore\vec v=\begin{pmatrix}u\cos\theta\\u\sin\theta-gt\end{pmatrix}

\therefore\vec r(t)=\begin{pmatrix}ut\cos\theta+c_3\\ut\sin\theta-\frac12gt^2+c_4\end{pmatrix}

\vec r(0)=\begin{pmatrix}0\\0\end{pmatrix}\Rightarrow c_3=c_4=0

\therefore\vec r(t)=\begin{pmatrix}ut\cos\theta\\ut\sin\theta-\frac12gt^2\end{pmatrix}

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Mechanics

Projectiles and resisted motionUse parametric and Cartesian equations to solve projectile problems with and without resistance.

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