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Vectors, EXT1 V1 EQ-Bank 2

A particle is projected from the origin at an angle \(\theta\) to the horizontal and an initial speed of \(V\) metres per second.

It passes through a point 37.5 metres above and 75 metres horizontally from its point of projection, as shown in the diagram.
 

Find the initial speed, \(V\), and angle of projection, \(\theta\), of the particle and determine the expression of the position vector of the particle, \(\underset{\sim}{r}(t)\), where \(t\) is the time after projection (use  \(g=10\ \text{ms}^{-2}\)).   (5 marks)

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\(\underset{\sim}{r}(t)=5 \sqrt{30} t \underset{\sim}{i}+\left(5 \sqrt{30} t-5 t^2\right) \underset{\sim}{j}\)

Show Worked Solution

\begin{array}{ll}
\dot{x}=V \cos \theta & \quad \dot{y}=V \sin \theta-10 t \\
x=V t \, \cos \theta & \quad y=V t \, \sin \theta-5 t^2
\end{array}

\(\text {Find} \ t \ \text{when}\ \ \dot{y}=0:\)

\(V \sin \theta-10 t=0 \ \  \Rightarrow \ \  t=\dfrac{V \sin \theta}{10}\)

\(\text{At max height,} \ \ \dot{y}=0 \ \ \text {and} \ \ x=\dfrac{1}{2} \times \ \text {range = 75 m}\)

\(\Rightarrow \ \text{At} \ \ t=\dfrac{V \sin \theta}{10}, x=75, y=37.5:\)

\(\dfrac{x}{y}\) \(=\dfrac{V t \cos \theta}{V t \sin \theta-5 t^2}\)  
\(\dfrac{75}{37.5}\) \(=\dfrac{V \cdot \dfrac{V \sin \theta}{10} \cdot \cos \theta}{V \cdot \dfrac{V \sin \theta}{10} \cdot \sin \theta-5\left(\dfrac{V \sin \theta}{10}\right)^2}\)  
\(2\) \(=\dfrac{V^2 \sin \theta \cos \theta}{10} \ ÷ \  \left(\dfrac{V^2 \sin ^2 \theta}{10}-\dfrac{V^2 \sin ^2 \theta}{20}\right)\)  
\(2\) \(=\dfrac{V^2 \sin \theta \cos \theta}{10} \times \dfrac{20}{V^2 \sin ^2 \theta}\)  
\(2\) \(=2 \cot \theta\)  
\(\theta\) \(=\tan ^{-1}(1)\)  
  \(=45^{\circ}\)  

 
 \(\text{Since} \ \ x=75 \ \ \text{when} \ \ t=\dfrac{V \sin 45°}{10}=\dfrac{V}{10 \sqrt{2}}:\)

\(75=V \cdot \dfrac{V}{10 \sqrt{2}} \cdot \dfrac{1}{\sqrt{2}}\)

\(V^2=1500\)

\(V=10 \sqrt{15} \quad(v>0)\)

\(\therefore \underset{\sim}{r}(t)\) \(=\dfrac{10 \sqrt{15}}{\sqrt{2}} t \underset{\sim}{i}+\left(\dfrac{10 \sqrt{15}}{\sqrt{2}} t-5 t^2\right) \underset{\sim}{j}\)
  \(=5 \sqrt{30} t \underset{\sim}{i}+\left(5 \sqrt{30} t-5 t^2\right) \underset{\sim}{j}\)

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

Show Worked Solution

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

  \(D(t)^2\) \(=(V t \cos \theta)^2+\left(V t \sin \theta-\dfrac{g t^2}{2}\right)^2\)
    \(=V^2 t^2 \cos ^2 \theta+V^2 t^2 \sin ^2 \theta-V g t^3 \sin \theta+\dfrac{g^2 t^4}{4}\)
    \(=V^2 t^2\left(\cos ^2 \theta+\sin ^2 \theta\right)-V g t^3 \sin \theta+\dfrac{g^2 t^4}{4}\)
    \(=V^2 t^2-V g t^3 \sin \theta+\dfrac{g^2 t^4}{4}\)
♦♦♦ 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 2019 SPEC2-N 4

A snowboarder at the Winter Olympics leaves a ski jump at an angle of `theta` degrees to the horizontal, rises up in the air, performs various tricks and then lands at a distance down a straight slope that makes an angle of 45° to the horizontal, as shown below.

Let the origin `O` of a cartesian coordinate system be at the point where the snowboarder leaves the jump, with a unit vector in the positive `x` direction being represented by  `underset~i`  and a unit vector in the positive `y` direction being represented by  `underset~j`. Distances are measured in metres and time is measured in seconds.

The position vector of the snowboarder  `t`  seconds after leaving the jump is given by

`underset~r (t) = (6t-0.01t^3) underset~i + (6 sqrt3 t-4.9t^2 + 0.01t^3) underset~j , \ t ≥ 0`
 


 

  1. Find the angle  `theta °`.    (2 marks)

  2. Find the speed, in metres per second, of the snowboarder when she leaves the jump at `O`.    (1 mark)

  3. Find the maximum height above `O` reached by the snow boarder. Give your answer in metres, correct to one decimal place.    (2 marks)

  4. Show that the time spent in the air by the snowboarder is  `(60(sqrt3 + 1))/(49)`  seconds.    (3 marks)

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  1. `60°`
  2. `12 \ text(ms)^-1`
  3. `5.5 \ text(m)`
  4. `text(See Worked Solutions)`
Show Worked Solution

a.   `v(t) = (6-0.03t^2)underset~i + (6 sqrt3-9.8t + 0.03t^2) underset~j`

`text(When) \ t =0,`

`v(t) = 6underset~i + 6 sqrt3 underset~j`

`tan theta = (6 sqrt3)/(6) = sqrt3`

`:. \ theta` `= tan^-1 sqrt3= 60°`

 

b.    `text(Speed)` `= |v(0)|`
    `= sqrt(6^2 + (6 sqrt3)^2)`
    `= 12 \ text(ms)^-1`

 
c.   `text(Max height when) \ underset~j \ text(component of) \ v(t) = 0`

`text(Solve): \ \ 6 sqrt3-9.8t + 0.03t^2 = 0`

`=> t =  1.064 \ text(seconds)`

`text(Max height)` `= 6 sqrt3 (1.064)-4.9(1.064)^2 + 0.01(1.064)^3`  
  `~~5.5\ text(m)`  

 
d.   `text(Time of Flight  ⇒  Solve for)\ \ t\ \ text(when)\ \ y=-x:`

`6 sqrt 3 t-4.9t^2 + 0.01 t^3` `= -(6t-0.01t^3)`
`(6 + 6 sqrt3)t-4.9 t^2` `= 0`
`t(6 + 6 sqrt3-4.9 t)` `= 0`
`4.9 t` `= 6 + 6 sqrt3`
`t` `= (6 + 6 sqrt3)/(4.9)`
  `= (60(sqrt3 + 1))/(49)\ text(seconds)`

Vectors, EXT1 V1 SM-Bank 6

A cricketer hits a ball at time  `t = 0`  seconds from an origin `O` at ground level across a level playing field.

The position vector  `underset ~s(t)`, from `O`, of the ball after `t` seconds is given by
 
  `qquad underset ~s(t) = 15t underset ~i + (15 sqrt 3 t - 4.9t^2)underset ~j`,
 
where,  `underset ~i`  is a unit vector in the forward direction, `underset ~j`  is a unit vector vertically up and displacement components are measured in metres.

  1. Find the initial velocity of the ball and the initial angle, in degrees, of its trajectory to the horizontal.  (2 marks)

  2. Find the maximum height reached by the ball, giving your answer in metres, correct to two decimal places.  (2 marks)

  3. Find the time of flight of the ball. Give your answer in seconds, correct to three decimal places.  (1 mark)

  4. Find the range of the ball in metres, correct to one decimal place.  (1 mark)

  5. A fielder, more than 40 m from `O`, catches the ball at a height of 2 m above the ground.

     

    How far horizontally from `O` is the fielder when the ball is caught? Give your answer in metres, correct to one decimal place.  (2 marks)

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  1. `underset ~v (0) = 15 underset ~i + 15 sqrt 3 underset ~j`

     

    `theta = pi/3 = 60^@`

  2. `34.44\ text(m)`
  3. `5.302\ text(s)`
  4. `79.5\ text(m)`
  5. `78.4\ text(m)`
Show Worked Solution

a.   `underset ~v (t) = underset ~ dot s (t) = 15 underset ~i + (15 sqrt 3 – 9.8t)underset ~j`

`text(Initial velocity occurs when)\ \ t=0:`

`:. underset ~v (0) = 15 underset ~i + 15 sqrt 3 underset ~j`
 

`text(Let)\ \ theta = text(Initial trajectory,)`

`tan theta` `=(15sqrt3)/15`  
  `=sqrt3`  
`:. theta` `=pi/3\ \ text(or)\ \ 60^@`  

 

b.  `text(Max height)\ =>underset~j\ \ text(component of)\ \ underset ~v=0.`

`15 sqrt 3 – 9.8t` `=0`
`t` `=(15 sqrt 3)/9.8`
  `=2.651…`

 
`text(Find max height when)\ \ t = 2.651…`

`:.\ text(Max height)` `= 15 sqrt 3 xx 2.651 – 4.9 xx (2.651)^2`
  `~~ 34.44\ text(m)`

 

c.   `text(Ball travels in symmetrical parabolic path.)`

`:.\ text(Total time of flight)`

`= 2 xx (15 sqrt 3)/9.8`

`= (15 sqrt 3)/4.9`

`~~ 5.302\ text(s)`
 

d.  `text(Range)\ =>underset~i\ \ text(component of)\ \ underset~s(t)\ \ text(when)\ \ t= (15 sqrt 3)/4.9`

`:.\ text(Range)` `= 15 xx (15 sqrt 3)/4.9`
  `= (225 sqrt 3)/4.9`
  `~~ 79.5\ text(m)`


e.   `text(Find)\ t\ text(when height of ball = 2 m:)`

`15 sqrt 3 t – 4.9t^2` `=2`  
`4.9t^2 – 15 sqrt 3 t + 2` `=0`  

 

  `t=(15 sqrt 3 +- sqrt ((15 sqrt 3)^2 – 4 xx 4.9 xx2))/(2 xx 4.9)`  
     

`t ~~ 0.078131\ \ text(or)\ \ t ~~ 5.22406`

 
`text(When)\ \ t=0.0781,`

`x= 15 xx 0.0781 = 1.17\ text(m)\ \ text{(no solution →}\ x<40 text{)}`
 

 `text(When)\ \ t=5.2241,`

`x=15 xx 5.2241 = 78.4\ text(m)`
 

`:.\ text(Ball is caught 78.4 m horizontally from)\ O.`