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Mechanics, EXT2* M1 2016 HSC 13b

The trajectory of a projectile fired with speed  `u\ text(ms)^-1`  at an angle  `theta`  to the horizontal is represented by the parametric equations

`x = utcostheta`   and   `y = utsintheta - 5t^2`,

where `t` is the time in seconds.

  1. Prove that the greatest height reached by the projectile is  `(u^2 sin^2 theta)/20`.  (2 marks)

A ball is thrown from a point `20\ text(m)` above the horizontal ground. It is thrown with speed `30\ text(ms)^-1` at an angle of `30^@` to the horizontal. At its highest point the ball hits a wall, as shown in the diagram.
 

     ext1-2016-hsc-q13
 

  1. Show that the ball hits the wall at a height of `125/4\ text(m)` above the ground.  (2 marks)

The ball then rebounds horizontally from the wall with speed `10\ text(ms)^-1`. You may assume that the acceleration due to gravity is `10\ text(ms)^-2`.

  1. How long does it take the ball to reach the ground after it rebounds from the wall?  (2 marks)

  2. How far from the wall is the ball when it hits the ground?  (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `2.5\ text(seconds)`
  4. `25\ text(m)`
Show Worked Solution
i.    `y` `= u t sin theta – 5t^2`
  `y prime` `= u sin theta – 10t`

 

`text(Maximum height when)\ \ y prime = 0`

`10 t` `= u sin theta`
`t` `= (u sin theta)/10`

 

`:.\ text(Maximum height)`

`= u ((u sin theta)/10) · sin theta – 5 ((u sin theta)/10)^2`

`= (u^2 sin^2 theta)/10 – (u^2 sin^2 theta)/20`

`= (u^2 sin^2 theta)/20\ text(… as required)`

 

ii.   `text{Using part (i)},`

`text(Height that ball hits wall)`

`= (30^2 · (sin 30)^2)/20 + 20`

`= (30^2 · (1/2)^2)/20 + 20`

`= 11 1/4 + 20`

`= 125/4\ text(m … as required)`

 

♦♦ Mean mark part (iii) 35%.
iii.   ext1-hsc-2016-13bi
`y ″` `= -10`
`y prime` `= -10 t`
`y` `= 125/4 – 5t^2`

 

`text(Ball hits ground when)\ \ y = 0,`

MARKER’S COMMENT: Many students struggled to solve: `5t^2=125/4`.
`5t^2` `= 125/4`
`t^2` `= 25/4`
`:. t` `= 5/2,\ \ t > 0`

 

`:.\ text(It takes the ball 2.5 seconds to hit the ground.)`

 

iv.   `text(Distance from wall)`

♦ Mean mark 43%.

`= 2.5 xx 10`

`= 25\ text(m)`

Mechanics, EXT2* M1 2007 HSC 7b

A small paintball is fired from the origin with initial velocity `14` metres per second towards an eight-metre high barrier. The origin is at ground level, `10` metres from the base of the barrier.

The equations of motion are

`x = 14t\ cos\ theta`

`y = 14t\ sin\ theta – 4.9t^2`

where  `theta`  is the angle to the horizontal at which the paintball is fired and  `t`  is the time in seconds. (Do NOT prove these equations of motion)
 

  1. Show that the equation of trajectory of the paintball is
     
         `y = mx − ((1 + m^2)/40)x^2`, where  `m = tan\ theta`.  (2 mark)

  2. Show that the paintball hits the barrier at height  `h`  metres when
     
         `m = 2 ± sqrt(3 − 0.4h)`.

     

    Hence determine the maximum value of  `h`.  (2 marks)

  3. There is a large hole in the barrier. The bottom of the hole is `3.9` metres above the ground and the top of the hole is `5.9` metres above the ground. The paintball passes through the hole if  `m`  is in one of two intervals. One interval is  `2.8 ≤ m ≤ 3.2`.

     

    Find the other interval.  (2 marks)

  4. Show that, if the paintball passes through the hole, the range is 
     
         `(40m)/(1 + m^2)\ \ text(metres.)`
     
    Hence find the widths of the two intervals in which the paintball can land at ground level on the other side of the barrier.  (3 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Proof (See Worked Solutions)}`
  3. `0.8 ≤ m ≤ 1.2`
  4. `text{1.3 m  (to 1 d.p.)  and  0.5 m  (to 1 d.p.)}`
Show Worked Solution
i.    `x` `= 14t\ cos\ theta` `\ \ …\ (1)`
  `y` `= 14t\ sin\ theta-4.9t^2` `\ \ …\ (2)`

 

`text(Substitute)\ \ t = x/(14\ cos\ theta)\ \ text{from (1) into (2)}`

`y` `= 14(x/(14\ cos\ theta))\ sin\ theta − 4.9(x/(14\ cos\ theta))^2`
  `= x\ tan\ theta − (4.9/(14^2))((x^2)/(cos^2\ theta))`
  `= x\ tan\ theta − (x^2)/40\ sec^2\ theta`
  `= x\ tan\ theta − (x^2)/40(1 + tan^2\ theta)`
  `= mx − ((1 + m^2)/40)x^2\ \ \ \ \ (text(Given)\ \ m = tan\ theta)`

 

ii.  `text(Show paintball hits at)\ \ h\ \ text(when)`

`m = 2 ± sqrt(3 − 0.4h)`

`text(i.e.)\ \ y = h\ \ text(when)\ \ x = 10`

`10m − ((1 + m^2)/40) · 10^2` `= h`
`10m − 5/2(1 + m^2)` `= h`
`20m − 5 − 5m^2` `= 2h`
`5m^2 − 20m + 2h + 5` `= 0`

 

`text(Using the quadratic formula)`

`m` `=(20 ± sqrt((-20)^2 − 4 · 5 · (2h + 5)))/(2 · 5)`
  `= (20 ± sqrt(400 − 40h −100))/10`
  `= (20 ± sqrt(300 − 40h))/10`
  `= (20 ± 10sqrt(3 − 0.4h))/10`
  `= 2 ± sqrt(3 − 0.4h)\ \ \ text(… as required)`

 

`text(Find maximum)\ \ h`

`sqrt(3 − 0.4h)` `≥ 0`
`3 − 0.4h` `≥ 0`
`0.4h` `≤ 3`
`h` `≤ 7.5`

 

`:.\ text(Maximum)\ \ h = 7.5\ text(m)`

 

iii.   EXT1 2007 7bi

`text{Using part (ii)}`

`text(When)\ \ h = 3.9`

`m` `= 2 ± sqrt(3 − 0.4(3.9))`
  `= 2 ± sqrt(1.44)`
  `= 2 ± 1.2`
  `= 3.2\ \ text(or)\ \ 0.8`

 

`text(When)\ \ h = 5.9`

`m` `= 2 ± sqrt(3 − 0.4(5.9))`
  `= 2 ± sqrt(0.64)`
  `= 2 ± 0.8`
  `= 2.8\ \ text(or)\ \ 1.2`

 

`:.\ text(The other interval is)\ \ \ 0.8 ≤ m ≤ 1.2`

 

iv.  `text(Find)\ \ x\ \ text(when)\ \ y = 0`

`mx − ((1 + m^2)/40)x^2` `= 0`
`x[m − ((1 + m^2)/40)x]` `= 0`
`((1 + m^2)/40)x` `= m,\ \  \ \ x ≠ 0`
`:. x` `= (40m)/(1 + m^2)\ \ …\ text(as required)`

 

`text(Consider the interval)\ \ \ 2.8 ≤ m ≤ 3.2`

`text(When)\ \ m = 2.8`

`=> x = (40(2.8))/(1 + 2.8^2) = 12.669…\ text(m)`

`text(When)\ \ m = 3.2`

`=>x = (40(3.2))/(1 + 3.2^2) = 11.387…\ text(m)`

 

`:.\ text(Landing width interval)`

`= 12.669… − 11.387…`

`= 1.281…`

`= 1.3\ text(m)\ \ text{to 1 d.p.}`

 

`text(Consider the interval)\ \ \ 0.8 ≤ m ≤ 1.2`

`text(When)\ \ m = 0.8`

`=>x = (40(0.8))/(1 + 0.8^2) = 19.512…\ text(m)`

`text(When)\ \ m = 1.2`

`=>x = (40(1.2))/(1 + 1.2^2) = 19.672…`

 

`text(S)text(ince interval includes)\ \ m = 1\ \ text(where the)`

`text(paintball has maximum range.)`

`x_(text(max)) = (40(1))/(1 + 1^2) = 20\ text(m)`

 

`:.\ text(Landing width interval)`

`= 20 − 19.512…`

`= 0.487…`

`= 0.5\ text(m)\ \ \ text{(to 1 d.p.)}`

Mechanics, EXT2* M1 2004 HSC 6b

A fire hose is at ground level on a horizontal plane. Water is projected from the hose. The angle of projection, `theta`, is allowed to vary. The speed of the water as it leaves the hose, `v` metres per second, remains constant. You may assume that if the origin is taken to be the point of projection, the path of the water is given by the parametric equations

`x = vt\ cos\ theta`

`y = vt\ sin\ theta − 1/2 g t^2`

where  `g\ text(ms)^(−2)`  is the acceleration due to gravity.  (Do NOT prove this.)

  1. Show that the water returns to ground level at a distance`(v^2\ sin\ 2theta)/g`  metres from the point of projection.   (2 marks)

This fire hose is now aimed at a 20 metre high thin wall from a point of projection at ground level 40 metres from the base of the wall. It is known that when the angle  `theta`  is 15°, the water just reaches the base of the wall.  
 

Calculus in the Physical World, EXT1 2004 HSC 6b
 

  1. Show that  `v^2 = 80g`.  (1 mark)

  2. Show that the cartesian equation of the path of the water is given by
     
         `y = x\ tan\ theta − (x^2\ sec^2\ theta)/160`.  (2 marks)

  3. Show that the water just clears the top of the wall if
     
         `tan^2\ theta − 4\ tan\ theta + 3 = 0`.  (2 marks)

  4. Find all values of  `theta`  for which the water hits the front of the wall.  (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
  3. `text(See Worked Solutions)`
  4. `text(See Worked Solutions)`
  5. `15^@ ≤ theta ≤ 45^@ \ \ text(and)\ \ 71.6^@ ≤ theta ≤ 75^@`
Show Worked Solution
i.    `x` `= vt\ cos\ theta`
  `y` `= vt\ sin\ theta − 1/2 g t^2`

 

`text(Find)\ \ t\ \ text(when)\ \ y = 0`

`vt\ sin\ theta − 1/2 g t^2` `= 0`
`t(v\ sin\ theta − 1/2 g t)` `= 0`
`v\ sin\ theta − 1/2 g t` `= 0, \ \ t ≠ 0`
`1/2 g t` `= v\ sin\ theta`
`t` `= (2v\ sin\ theta)/g`

 

`text(Find)\ \ x\ \ text(when)\ \ t = (2v\ sin\ theta)/g`

`x` `= v · (2v\ sin\ theta)/g\ cos\ theta`
  `= (v^2 · \ 2\ sin\ theta\ cos\ theta)/g`
  `= (v^2\ sin\ 2theta)/g`

 

`:.\ text(When)\ \ x = (v^2\ sin\ 2theta)/g,\ \ \text(the water returns)`

`text(to ground level  … as required.)`

 

ii.   `text(Show)\ \ v^2 = 80\ text(g)`

`text(When)\ \ theta = 15^@, \ x = 40`

`text{From part (i)}`

`40` `= (v^2\ sin\ 30^@)/g`
`v^2 xx 1/2` `= 40g`
`v^2` `= 80g\ \ \ …text(as required)`

 

iii.  `text(Show)\ \ y = x\ tan\ theta − (x^2\ sec^2\ theta)/160`

`x` `= vt\ cos\ theta`
`:.t` `= x/(v\ cos\ theta)`

 

`text(Substitute into)`

`y` `= vt\ sin\ theta − 1/2 g t^2`
  `= v · x/(v\ cos\ theta) · sin\ theta −1/2 g  (x/(v\ cos\ theta))^2`
  `= x\ tan\ theta − 1/2  g (x^2/(v^2\ cos^2\ theta))`
  `= x\ tan\ theta − 1/2  g · x^2/(80g\ cos^2\ theta)`
  `= x\ tan\ theta − (x^2\ sec^2\ theta)/160\ \ \ …text(as required.)`

 

iv.   `text(Water clears the top if)\ \ y = 20\ \ text(when)\ \ x = 40`

`text{Substitute into equation from (iii)}`

`40\ tan\ theta − (40^2\ sec^2\ theta)/160` `= 20`
`40\ tan\ theta − 10\ sec^2\ theta` `= 20`
`40\ tan\ theta − 10(1 + tan^2\ theta)` `= 20`
`40\ tan\ theta − 10 − 10\ tan^2\ theta` `= 20`
`10\ tan^2\ theta − 40\ tan\ theta\ + 30` `= 0`
`tan^2\ theta − 4\ tan\ theta + 3` `= 0\ \ \ text(… as required)`

 

v.   

Calculus in the Physical World, EXT1 2004 HSC 6b Answer

`text(Water hits the bottom of the wall when)`

`x = 40\ \ text(and)\ \ y = 0`

`40\ tan\ theta − (40^2\ sec^2\ theta)/160` `= 0`
`40\ tan\ theta − 10\ sec^2\ theta` `= 0`
`4\ tan\ theta − (1 + tan^2\ theta)` `= 0`
`tan^2\ theta − 4\ tan\ theta + 1` `= 0`

 

`text(Using the quadratic formula)`

`tan\ theta` `= (+4 ± sqrt(16 − 4 · 1 · 1))/ (2 xx 1)`
  `= (4 ± sqrt12)/2`
  `= 2 ± sqrt3`
`theta` `= 15^@\ \ text(or)\ \ 75^@`

 

`text(Water hits the top of the wall when)`

`x = 40\ text(and)\ \ y = 20`

`tan^2\ theta − 4\ tan\ theta + 3` `= 0\ \ \ text{from (iv)}`
`(tan\ theta − 1)(tan\ theta − 3)` `= 0`
`tan\ theta` `= 1` `text(or)` `tan\ theta` `= 3`
`theta` `= 45^@`   `theta` `= tan^(−1)\ 3`
        `= 71.565…`
        `= 71.6^@\ \ \text{(to 1 d.p.)}`

 

`:.\ text(Following the motion of the water as)\ \ theta\ \ text(increases,)`

`text(water hits the wall when)`

`15^@ ≤ theta ≤ 45^@` `\ text(and)`
`71.6^@ ≤ theta ≤ 75^@`