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Vectors, EXT1 V1 2024 HSC 14d

A particle is projected from the origin, with initial speed \(V\) at an angle of \(\theta\) to the horizontal. The position vector of the particle, \(\underset{\sim}{r}(t)\), where \(t\) is the time after projection and \(g\) is the acceleration due to gravity, is given by

\(\underset{\sim}{r}(t)=\left(\begin{array}{c}Vt\cos\theta \\Vt\sin \theta -\dfrac{gt^2}{2}\end{array}\right)\).   (Do NOT prove this.)

Let \(D(t)\) be the distance of the particle from the origin at time \(t\), so  \(D(t)=|\underset{\sim}{r}(t)|\).

Show that for  \(\theta<\sin ^{-1}\left(\sqrt{\dfrac{8}{9}}\right)\)  the distance, \(D(t)\), is increasing for all  \(t>0\).   (4 marks)

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\(\text{See Worked Solutions.}\)

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\(\underset{\sim}{r}(t)=\left(\begin{array}{c}Vt\cos\theta \\Vt\sin \theta -\dfrac{gt^2}{2}\end{array}\right), \ D(t)=\abs{\underset{\sim}{r}(t)}\)

  \(D(t)^2\) \(=(V t \cos \theta)^2+\left(V t \sin \theta-\dfrac{g t^2}{2}\right)^2\)
    \(=V^2 t^2 \cos ^2 \theta+V^2 t^2 \sin ^2 \theta-V g t^3 \sin \theta+\dfrac{g^2 t^4}{4}\)
    \(=V^2 t^2\left(\cos ^2 \theta+\sin ^2 \theta\right)-V g t^3 \sin \theta+\dfrac{g^2 t^4}{4}\)
    \(=V^2 t^2-V g t^3 \sin \theta+\dfrac{g^2 t^4}{4}\)
♦♦♦ Mean mark 25%.

\(\text{Let }F(t)=D(t)^2\)

  \(F^{\prime}(t)\) \(=2 V^2 t-3 V g t^2 \sin \theta+g^2 t^3\)
    \(=t\left(2 V^2-3 V g t \sin \theta+g^2 t^2\right)\)

\(\text {Monotonically increasing when } F^{\prime}(t)>0 \text { for all } t.\)

\(g^2 t^2-3 V g t\, \sin \theta+2 V^2>0\)
 

\(\text {Find } t \text { when } \Delta<0:\)

  \((-3 V g \sin \theta)^2-4 \cdot g^2 \cdot 2 V^2\) \(<0\)
  \(9 V^2 g^2 \sin ^2 \theta-8 V^2 g^2\) \(<0\)
  \(\sin ^2 \theta\) \(<\dfrac{8}{9}\)
  \(\sin \theta\) \(<\sqrt{\dfrac{8}{9}} \ \ \left(\theta \in\left(0, \dfrac{\pi}{2}\right) \Rightarrow \ \sin \theta \in(0,1)\right)\)
  \(\theta\) \(<\sin ^{-1}\left(\sqrt{\dfrac{8}{9}}\right)\)

Vectors, EXT1 V1 SM-Bank 30

A canon ball is fired from a castle wall across a horizontal plane at `V` ms−1.

Its position vector  `t` seconds after it is fired from its origin is given by  `underset~s(t) = V tunderset~i - 1/2g t^2 underset~j`.

  1. If the projectile hits the ground at a distance 8 times the height at which it was fired, show that it initial velocity is given by
     
          `V = 4sqrt(2hg)`  (2 marks)

  2. Show that the total distance the canon ball travels can be expressed as
     
          `int_0^sqrt((2h)/g) sqrt(g(32h + g t^2))\ dt`      (2 marks)

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  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
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i.   `text(Time of flight  ⇒  find)\ t\ text(when)\ \ y = −h`

`−1/2g t^2` `= −h`
`t^2` `= (2h)/g`
`t` `= sqrt((2h)/g),\ \ t > 0`

 

`text(S)text(ince the canon ball impacts when)\ \ x = 8h:`

`Vt` `= 8h`
`Vsqrt((2h)/g)` `= 8h`
`V` `= (8sqrt(hg))/sqrt2`
  `= 4sqrt(2hg)`

 

ii.   `underset~v = 4sqrt(2hg) underset~i – g t underset~j`

`|underset~v|` `= sqrt((4sqrt(2hg))^2 + (−g t)^2)`
  `= sqrt(16 xx 2hg  + g^2 t^2)`
  `=sqrt(g(32h + g t^2)`

 

`text(Distance)` `= int_0^sqrt((2h)/g) |underset~v|\ dt`
  `= int_0^sqrt((2h)/g) sqrt(g(32h + g t^2))\ dt`