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Vectors, EXT1 V1 2025 HSC 14b

Points \(A\) and \(B\) lie vertically above the origin. Point \(A\) is higher than point \(B\) such that  \(\dfrac{OA}{O B}=k\), where  \(k>1\).

A particle is projected horizontally from point \(A\) with velocity \(U\ \text{ms}^{-1}\). After \(T\) seconds, another particle is projected horizontally from point \(B\) with velocity \(V\ \text{ms}^{-1}\). The two particles land on the ground in the same place.
 

Show that the ratio \(\dfrac{V}{U}\) depends only on \(k\).   (4 marks)

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

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\(\text{Let} \ \ OB=h \ \Rightarrow \ OA=k \times OB=k h\)

\({\underset{\sim}{a}}_A=\displaystyle \binom{0}{-g}, \quad {\underset{\sim}{v}}_A=\displaystyle \binom{U}{-g t_1}, \quad {\underset{\sim}{r}}_A=\displaystyle \binom{U t_1}{k h-\frac{1}{2} g t_1^2}\)

 
\(\text{Time of flight for} \ B\left(t_2\right) \neq \text{Time of flight for} \ A\left(t_1\right)\)

\({\underset{\sim}{a}}_B=\displaystyle \binom{0}{-g}, \quad {\underset{\sim}{v}}_B=\displaystyle\binom{V}{-g t_2}, \quad {\underset{\sim}{r}}_B=\displaystyle\binom{V t_2}{h-\frac{1}{2} g t_2^2}\)
 

\(\text{Time of flight for} \ A\left(t_1\right)\) :

\(kh-\dfrac{1}{2} g t_1^2=0 \ \Rightarrow \  t_1^2=\dfrac{2kh}{g} \ \Rightarrow \ t_1=\sqrt{\dfrac{2 kh}{g}}\)

\(\text{Range of} \ A=Ut_1=U \sqrt{\dfrac{2 k h}{g}}\)

 
\(\text{Time of flight for}\ B\left(t_2\right):\)

\(h-\dfrac{1}{2} g t_2^2=0 \ \Rightarrow \ t_2^2=\dfrac{2 h}{g}\ \Rightarrow \  t_2=\sqrt{\dfrac{2 h}{g}}\)

\(\text{Range of} \ B=Vt_2=V \sqrt{\dfrac{2h}{g}}\)

\(\text{Equating ranges:}\)

\(V \sqrt{\dfrac{2h}{g}}\) \(=U \sqrt{\dfrac{2 kh}{g}}\)  
\(\dfrac{V}{U}\) \(=\sqrt{\dfrac{2 kh}{g}} \times \sqrt{\dfrac{g}{2 h}}=\sqrt{k}\)  

 
\(\therefore \dfrac{V}{U} \ \text{ratio depends only on} \ k.\)

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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`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text{Projectile’s motion:}`

`x_P=2utcostheta, \ \ y_P=2utsintheta-g/2t^2`

`text{Time of flight}\ \ =>\ \ text{Find}\ \ t\ \ text{when}\ \ y=0:`

`2utsintheta-g/2t^2` `=0`  
`t(2usintheta-g/2 t)` `=0`  
`g/2 t` `=2u sin theta`  
`t` `=(4usintheta)/g`  

 

`text{Range of projectile}\ (R):`

`R` `=2u((4usintheta)/g) costheta`  
  `=(8u^2sinthetacostheta)/g`  
  `=(4u^2sin(2theta))/g`  

 
`=>R_max=(4u^2)/g \ \ (text{when}\ \ theta=45°):`

`text{Projectile will hit target when}\ \ x_P=x_T\ \ text{at}\ \ t=(4usintheta)/g:`

`(4u^2sin(2theta))/g` `=d+u*(4usintheta)/g`  
`d` `=(4u^2sin(2theta))/g-u*(4usintheta)/g`  
  `=(4u^2)/g\ (sin(2theta)-sintheta)`  
  `=R_max*(sin(2theta)-sintheta)`  

 
`d_max\ \ text{occurs when}\ \ (sin(2theta)-sin theta)\ \ text{is MAX}`

`f(theta)` `=sin(2theta)-sintheta`  
`f^(′)(theta)` `=2cos(2theta)-costheta`  
`f^(′′)(theta)` `=-4sin(2theta)+sin theta`  

 
`text{Find SP’s when}\ \ f^(′)(theta)=0:`

`2cos2theta-costheta` `=0`  
`2(2cos^2theta-1)-costheta` `=0`  
`4cos^2theta-costheta-2` `=0`  

 

`costheta` `=(1+-sqrt(1-4xx4xx-2))/(2xx4)`  
  `=(1+-sqrt33)/8`  
`theta` `=32°32′, \ 126°23′`  

 
`f^(′′)(32°32′)~~-3.1<0`

`=>\ text{MAX at}\ \ theta=32°32′`
 

`:.d_max` `=R_max xx (sin65°04′-sin32°32′)`  
  `=(0.369)R_max<0.37R_max`  

♦♦ Mean mark 33%.

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)

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`text(See Worked Solution)`

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`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 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)`
Show Worked Solution

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`

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)

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  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 SM-Bank 9

The diagram shows a projectile fired at an angle  `theta`  to the horizontal from the origin `O` with initial velocity  `V\ text(ms)^(−1)`.
 

The position vector for the projectile is given by
 

`qquad underset~s(t) = Vtcosthetaunderset~i + (Vtsintheta - 1/2 g t^2)underset~j`     (DO NOT prove this)
 

where `g` is the acceleration due to gravity.

  1.  Show the horizontal range of the projectile is

    `qquad (V^2sin2theta)/g`  (2 marks)

The projectile is fired so that  `theta = pi/3`.

  1.  State whether the projectile is travelling upwards or downwards when

    `qquad t = (2V)/(sqrt3g)`  (1 mark)

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  1. `text(See Worked Solutions)`
  2. `text(Downwards – See Worked Solutions)`
Show Worked Solution

i.   `text(Time of flight when)`

`underset~j\ text(component of)\ underset~v = 0`

`Vtsintheta – 1/2 g t^2` `= 0`
`t(Vsintheta – 1/2 g t)` `= 0`
`1/2g t` `= Vsintheta`
`t` `= (2Vsintheta)/g`

 
`text(Range) \ => \ underset~i\ text(component of)\ underset~s`

`text(when) \ \ t = (2Vsintheta)/g`

`text(Range)` `= V · ((2Vsintheta)/g) · costheta`
  `= (V^2)/g · 2sinthetacostheta`
  `= (V^2sin2theta)/g`

 

ii.   `text(Time of flight) = (2Vsin\ pi/3)/g = (sqrt3 V)/g`

`text(S)text(ince parabolic path is symmetrical,)`

`=>\ text(Upwards if)\ \ t < (sqrt3 V)/(2g)`

`=>\ text(Downwards if)\ \ t > (sqrt3 V)/(2g)`

`:. \ text(At)\ \ t = (2V)/(sqrt3 g), text(travelling downwards)`

`text(as) \ \ 2/sqrt3 · V/g > sqrt3/2 · V/g`

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.`