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Vectors, EXT1 V1 2023 HSC 13b

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.

  1. By first showing that  \(\cos \theta=\dfrac{1}{\sqrt{5}}\), verify that  \(v=\sqrt{5} u\).   (2 marks)

  2. Show that the particles collide at time  \(T=\dfrac{H}{2 u}\).   (1 mark)

When the particles collide, their velocity vectors are perpendicular.

  1. Show that  \(H=\dfrac{2 u^2}{g}\).   (3 marks)

  2. Prior to the collision, the trajectory of particle \(A\) was a parabola. (Do NOT prove this.)  
  3. Find the height of the vertex of that parabola above the horizontal plane. Give your answer in terms of \(H\).   (2 marks)

Show Answers Only

  1. \(\text{See Worked Solutions}\)
  2. \(\text{See Worked Solutions}\)
  3. \(\text{See Worked Solutions}\)
  4. \(H\)

Show Worked Solution

i.    \(\text{Given}\ \ \tan \theta =2\)

\(\cos \theta = \dfrac{1}{\sqrt 5} \)

\(\text{Since particles collide, for some}\ t: \)

\(vt\ \cos \theta\) \(=ut\)  
\(v \cdot \dfrac{1}{\sqrt 5} \) \(=u\)  
\(v\) \(=\sqrt5 u\)  


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.  \(\text{Velocity vectors:} \)

\[\textbf{v}_A(t)=\left(\begin{array}{c}v\ \cos \theta \\v\ \sin\theta-gt\end{array}\right)=\left(\begin{array}{c} u \\ 2u-gt \end{array}\right)\]

\[\textbf{v}_B(t)=\left(\begin{array}{c}u \\-gt \end{array}\right)\]

\(\text{Since particles are perpendicular at collision:}\)

\(\textbf{v}_A \cdot \textbf{v}_B=0\)

♦♦ Mean mark (iii) 37%.
\(u^2+(-gt)(2u-gt)\) \(=0\)  
\(u^2-2gtu+g^2t^2\) \(=0\)  
\((u-gt)^2\) \(=0\)  
\(gt\) \(=u\)  
\(t\) \(=\dfrac{u}{g}\)  
\(\dfrac{H}{2u}\) \(=\dfrac{u}{g}\ \ \text{(see part (ii))}\)  
\(\therefore H\) \(=\dfrac{2u^2}{g}\)  

 
iv.   \(\text{Height of vertex}\ \ \Rightarrow \ \text{Find}\ t\ \text{when y-component of}\ \textbf{v}_A=0 \)

\(v\ \sin \theta-gt\) \(=0\)  
\(t\) \(=\dfrac{v\ \sin \theta}{g} \)  

 
\(\text{Height of vertex}\ =\ \text{y-component of}\ \textbf{r}_A\ \text{when}\ \ t= \dfrac{v\ \sin \theta}{g} \)

\(\text{Height}\) \(=vt\ \sin \theta-\dfrac{1}{2}gt^2 \)  
  \(=\dfrac{v^2\sin^2 \theta}{g}-\dfrac{1}{2}g\Big{(} \dfrac{v^2\sin^2 \theta}{g^2}\Big{)} \)  
  \(=\dfrac{v^2 \sin^2 \theta}{2g} \)  
  \(=\dfrac{(2u)^2}{2g} \)  
  \(=\dfrac{2u^2}{g} \)  
  \(=H\)  
♦♦ Mean mark (iv) 34%.

Vectors, EXT1 V1 2021 HSC 13b

When an object is projected from a point `h` metres above the origin with initial speed `V` m/s at an angle of  `theta^@`  to the horizontal, its displacement vector, `t` seconds after projection, is

`underset~r(t) = (Vtcostheta)underset~i + (-5t^2 + Vtsintheta + h)underset~j`.     (Do NOT prove this.)

A person, standing in an empty room which is 3 m high, throws a ball at the far wall of the room. The ball leaves their hand 1 m above the floor and 10 m from the far wall. The initial velocity of the ball is 12 m/s at an angle of 30° to the horizontal.

Show that the ball will NOT hit the ceiling of the room but that it will hit the far wall without hitting the floor.  (4 marks)

Show Answers Only

`text(See Worked Solution)`

Show Worked Solution


 

`r(t) = (Vtcos theta)underset~i + (-5t^2 + Vtsintheta + h)underset~j`

`r′(t) = (Vcostheta)underset~i + (-10t + Vsintheta)underset~j`

`text(Max height occurs when)\ \ dot y = 0:`

`10t` `= Vsintheta`
`t` `= (12 xx sin30^@)/10`
  `= 0.6\ text(sec)`

 
`text(Find)\ \ y\ \ text(when)\ \ t = 0.6:`

`y` `= -5(0.6)^2 + 12 xx 0.6 xx sin30^@ + 1`
  `= 2.8\ text(m < 3 m)`

 
`:.\ text(Ball will not hit ceiling.)`
 

`text(Find time of flight when)\ \  y = 0:`

`-5t^2 + 6t + 1` `= 0`
`5t^2 – 6t – 1` `= 0`
`t` `= (6 +- sqrt((-6)^2 + 4 · 5 · 1))/10`
  `= (6 + sqrt56)/10`
  `= 1.348…`

 
`text(Find)\ \ x\ \ text(when)\ \ t = 1.348:`

`x` `= 12 xx 1.348 xx cos 30`
  `= 14.0 > 10`

 
`:.\ text(Ball will hit the wall on the full.)`

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)

Show Answers Only
  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 23

A fireworks rocket is fired from an origin `O`, with a velocity of 140 metres per second at an angle of  `theta`  to the horizontal plane.
 


 

The position vector `underset~s(t)`, from `O`, of the rocket after  `t`  seconds is given by

`underset~s = 140tcosthetaunderset~i + (140tsintheta - 4.9t^2)underset~j`

The rocket explodes when it reaches its maximum height.

  1. Show the rocket explodes at a height of  `1000sin^2theta`  metres.  (2 marks)

  2. Show the rocket explodes at a horizontal distance of  `1000sin 2theta`  metres from `O`.  (1 mark)

  3. For best viewing, the rocket must explode at a horizontal distance of 500 m and 800 m from `O`, and at least 600 m above the ground.

     

    For what values of  `theta`  will this occur.  (3 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `63.4° <= theta <= 75°`
Show Worked Solution

i.    `underset~s = 140tcosthetaunderset~i + (140tsintheta – 4.9t^2)underset~j`

`underset~v = 140costhetaunderset~i + (140sintheta – 9.8t)underset~j`

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

`0` `= 140sintheta – 9.8t`
`t` `= (140sintheta)/9.8`

 
`text(Max height:)\ \ underset~j\ text(component of)\ underset~s\ text(when)\ t = (140sintheta)/9.8`

`text(Max height)` `= 140sintheta · (140sintheta)/9.8 – (4.9 · 140^2sin^2theta)/(9.8^2)`
  `= 2000sin^2theta – 1000sin^2theta`
  `= 1000sin^2theta`

 

ii.   `text(Horizontal distance)\ (d):`

`=>\ underset~i\ text(component of)\ underset~s\ text(when)\ \ t = (140sintheta)/9.8`

`:.d` `= 140costheta · (140sintheta)/9.8`
  `= (140 xx 70 xx sin2theta)/9.8`
  `= 1000sin2theta`

 

iii.   `text(Using part ii),`

`500<=1000sin2theta<=800`
`0.5<=sin2theta<=0.8`

 

`text(In the 1st quadrant:)`

`30° <=` `2theta` `<= 53.13°`
`15° <=` `theta` `<= 26.6°`

 
`text(In the 2nd quadrant:)`

`126.87°<=` `2theta` `<= 150°`
`63.4°<=` `theta` `<= 75°`

 
`text(When)\ theta = 26.6°:`

`text(Max height)` `= 1000 · sin^2 26.6°`
  `= 200.5\ text(metres)\ (< 600\ text(m))`

 
`=>\ text(Highest max height for)\ \ 15° <= theta < 26.6°\ \ text(does not satisfy.)`
 

`text(When)\ theta = 63.4°:`

`text(Max height)` `= 1000 · sin^2 63.4°`
  `= 799.5\ text(metres)\ (> 600\ text(m))`

 
`=>\ text(Lowest max height for)\ \ 63.4° <= theta <= 75°\ \ text(satisfies).`

`:. 63.4° <= theta <= 75°`

Vectors, EXT1 V1 EQ-Bank 1

A basketball player aims to throw a basketball through a ring, the centre of which is at a horizontal distance of 4.5 m from the point of release of the ball and 3 m above floor level. The ball is released at a height of 1.75 m above floor level, at an angle of projection `alpha` to the horizontal and at a speed of  `V\ text(ms)^(-1)`. Air resistance is assumed to be negligible.
 


 

The position vector of the centre of the ball at any time, `t` seconds, for  `t >= 0`, relative to the point of release is given by 
 
`qquad underset ~s(t) = Vt cos (alpha) underset ~i + (Vt sin(alpha) - 4.9t^2) underset ~j`,
 
where  `underset ~i`  is a unit vector in the horizontal direction of motion of the ball and  `underset ~j`  is a unit vector vertically up. Displacement components are measured in metres.

For the player’s first shot at goal, `V = 7\ text(ms)^(-1)` and  `alpha = 45^@`

  1. Find the time, in seconds, taken for the ball to reach its maximum height. Give your answer in the form  `(a sqrt b)/c`, where  `a, b` and `c` are positive integers.  (2 marks) 

  2. Find the maximum height, in metres, above floor level, reached by the centre of the ball.  (2 marks)

  3. Find the distance of the centre of the ball from the centre of the ring one second after release. Give your answer in metres, correct to two decimal places.  (2 marks)

Show Answers Only
  1. `(5 sqrt 2)/14`
  2. `3\ text(m)`
  3. `128\ text(m)`
Show Worked Solution
a.i.    `underset ~s(t)` `=7t\ cos 45^@ underset ~i+(7t\ sin 45^@ – 4.9t^2) underset ~j`
    `=(7sqrt2 t)/2 underset ~i + ((7sqrt2)/2 t – 4.9t^2) underset ~j`
     

 `text(Maximum height  ⇒  find)\ \t\ text(when)\ \ underset~j\ \ text(component of)\ \ underset ~V(t) = 0:`

`underset~V = (7sqrt2)/2 underset ~i + ((7sqrt2)/2 – 9.8t) underset ~j`

`(7sqrt2)/2 – 9.8t` `= 0`
`t` `= (5 sqrt 2)/14\ \ text(seconds)`

 

a.ii.  `text(Max height)\ =>\ text(Find)\ \ underset~j\ \ text(component of)\ \ underset~s(t)\ \ text(when)\ \ t=(5 sqrt 2)/14`

  `underset ~s_(underset ~j)((5 sqrt 2)/14)` `= 7 xx (5 sqrt 2)/14 xx 1/sqrt 2 – 4.9 xx ((5 sqrt 2)/14)^2`
    `= 1.25\ text(m)`

 
`:.\ text(Height above floor) = 1.25 + 1.75 = 3\ text(m)`

 

a.iii.  `underset ~s_text(ring)= 4.5 underset ~i + 1.25 underset ~j`

`text(After 1 second,)`

`underset~s_text(ball) = 7/sqrt 2 underset ~i + ((7sqrt2)/2 – 4.9)underset ~j`

  `:.d` `= |underset ~s_text(ring) – underset ~s(1)|`
    `= sqrt((4.5 – 7/sqrt 2)^2 + (1.25 – 7/sqrt 2 + 4.9)^2)`
    `~~ 1.28\ text(m)`

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)

Show Answers Only
  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.`