Particle \(A\) is projected from the origin with initial speed \(v\) m s\(^{-1}\) at an angle \(\theta\) with the horizontal plane. At the same time, particle \(B\) is projected horizontally with initial speed \(u\) ms\(^{-1}\) from a point that is \(H\) metres above the origin, as shown in the diagram. The position vector of particle \(A, t\) seconds after it is projected, is given by \[\textbf{r}_A(t)=\left(\begin{array}{c}v t\ \cos \theta \\vt\ \sin\theta-\dfrac{1}{2} g t^2\end{array}\right) \text{. (Do NOT prove this.)}\] The position vector of particle \(B, t\) seconds after it is projected, is given by \[\textbf{r}_B(t)=\left(\begin{array}{c}u t \\H-\dfrac{1}{2} g t^2\end{array}\right) \text{. (Do NOT prove this.)}\] The angle \(\theta\) is chosen so that \(\tan \theta=2\). The two particles collide. --- 5 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- When the particles collide, their velocity vectors are perpendicular. --- 6 WORK AREA LINES (style=lined) --- --- 5 WORK AREA LINES (style=lined) ---
i. \(\text{Given}\ \ \tan \theta =2\) \(\cos \theta = \dfrac{1}{\sqrt 5} \) \(\text{Since particles collide, for some}\ t: \) \[\textbf{v}_A(t)=\left(\begin{array}{c}v\ \cos \theta \\v\ \sin\theta-gt\end{array}\right)=\left(\begin{array}{c} u \\ 2u-gt \end{array}\right)\] \[\textbf{v}_B(t)=\left(\begin{array}{c}u \\-gt \end{array}\right)\]
\(\text{Since particles are perpendicular at collision:}\) \(\textbf{v}_A \cdot \textbf{v}_B=0\)
\(vt\ \cos \theta\)
\(=ut\)
\(v \cdot \dfrac{1}{\sqrt 5} \)
\(=u\)
\(v\)
\(=\sqrt5 u\)
ii. \(\text{Equating y-components of}\ \textbf{r}_A\ \text{and}\ \textbf{r}_B : \)
\(vt\ \sin \theta-\dfrac{1}{2}gt^2\)
\(=H-\dfrac{1}{2}gt^2\)
\(vt\ \sin \theta\)
\(=H\)
\(u \sqrt{5} \times t \times \dfrac{2}{\sqrt5} \)
\(=H\)
\(t\)
\(= \dfrac{H}{2u} \)
iii. \(\text{Velocity vectors:} \)
\(u^2+(-gt)(2u-gt)\)
\(=0\)
\(u^2-2gtu+g^2t^2\)
\(=0\)
\((u-gt)^2\)
\(=0\)
\(gt\)
\(=u\)
\(t\)
\(=\dfrac{u}{g}\)
\(\dfrac{H}{2u}\)
\(=\dfrac{u}{g}\ \ \text{(see part (ii))}\)
\(\therefore H\)
\(=\dfrac{2u^2}{g}\)
iv. \(\text{Height of vertex}\ \ \Rightarrow \ \text{Find}\ t\ \text{when y-component of}\ \textbf{v}_A=0 \)
\(v\ \sin \theta-gt\)
\(=0\)
\(t\)
\(=\dfrac{v\ \sin \theta}{g} \)
\(\text{Height of vertex}\ =\ \text{y-component of}\ \textbf{r}_A\ \text{when}\ \ t= \dfrac{v\ \sin \theta}{g} \)
\(\text{Height}\)
\(=vt\ \sin \theta-\dfrac{1}{2}gt^2 \)
\(=\dfrac{v^2\sin^2 \theta}{g}-\dfrac{1}{2}g\Big{(} \dfrac{v^2\sin^2 \theta}{g^2}\Big{)} \)
\(=\dfrac{v^2 \sin^2 \theta}{2g} \)
\(=\dfrac{(2u)^2}{2g} \)
\(=\dfrac{2u^2}{g} \)
\(=H\)
Vectors, EXT1 V1 2022 HSC 14c
A video game designer wants to include an obstacle in the game they are developing. The player will reach one side of a pit and must shoot a projectile to hit a target on the other side of the pit in order to be able to cross. However, the instant the player shoots, the target begins to move away from the player at a constant speed that is half the initial speed of the projectile shot by the player, as shown in the diagram below.
The initial distance between the player and the target is `d`, the initial speed of the projectile is `2 u` and it is launched at an angle of `theta` to the horizontal. The acceleration due to gravity is `g`. The launch angle is the ONLY parameter that the player can change.
Taking the position of the player when the projectile is launched as the origin, the positions of the projectile and target at time `t` after the projectile is launched are as follows.
| `vecr_(P)` | `=((2utcostheta),(2utsintheta-g/2t^2))` | `text{Projectile}` |
| `vecr_(T)` | `=((d+ut),(0))` | `text{Target (Do NOT prove these)}` |
Show that, for the player to have a chance of hitting the target, `d` must be less than 37% of the maximum possible range of the projectile (to 2 significant figures). (4 marks)
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