SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Mechanics, EXT2 M1 2021 HSC 16b

A particle which is projected from the origin with initial speed `u` ms-1 at an angle of `theta` to the positive `x`-axis lands on the `x`-axis, as shown in the diagram. The particle is subject to an acceleration due to gravity of `g` ms-1. 
 

The position vector of the particle, `underset~r (t)`, where `t` is the time in seconds after the particle is projected, is given by

`underset~r (t) = ((ut cos theta),( - {g t^2}/{2} + u t sin theta))`.     (Do NOT prove this.)

For some value(s) of  `theta`  there will be two times during the time of flight when the particle’s position vector is perpendicular to its velocity vector.

Find the value(s) of  `theta`  for which this occurs, justifying that both times occur during the time of flight.  (5 marks)

Show Answers Only

`text{See Worked Solutions}`

Show Worked Solution
`underset~r (t) = ((ut cos theta),( – {g t^2}/{2} + u t sin theta)) , \ underset~v (t) = ((u cos theta),(-g t + u sin theta))`
♦♦ Mean mark 31%.
 
`text{If} \ \ underset~r (t) ⊥ underset~v (t) \ => \ underset~r (t) * underset~v (t) = 0`
 
`u t cos theta * u cos theta + (-{g t^2}/{2} + u t sin theta) (- g t + u sin theta)` `= 0`
`u^2 t cos^2 theta + {g^2 t^3}/{2} – {g t^2}/{2} u sin theta – g t^2 u sin theta + u^2 t sin theta` `= 0`
`u^2 t (cos^2 theta + sin^2 theta) + {g^2 t^3}/{2} – {3 u g t^2}/{2} sin theta` `= 0`
`u^2 t + {g^2 t^3}/{2} – {3 u g t}/{2} sin theta` `= 0`
`g^2 t^2 – 3 u g t sin theta + 2 u^2` `= 0`

 

`text{If 2 roots} , \ Δ > 0 :`

`(-3 u g sin theta)^2 – 4 g^2 \ 2 u^2` `> 0`
`9 u^2 g^2 sin^2 theta -8 g^2 u^2` `> 0`
`u^2 g^2 (9 sin^2 theta – 8)` `> 0`
`sin^2 theta` `> 8/9`
`theta` `> sin^-1 ({2 sqrt2}/{3})`

 

`text{S} text{ince} \ \ 0 < theta < pi/2 \ , text{only one value of} \ theta \ text{satisfies}`

`text{Checking valid times of flight}\ (t_f):`

`1/2 \ text{time of flight} \ => underset~j text{-component of} \ underset~v (t) = 0`

`1/2 g t_f` `= u sin theta`
`t_f` `= {2 u sin theta}/{g}`

 

`t` `={3 u g sin theta ± sqrt{9 u^2 g^2 sin^2 theta – 4g^2 * 2 u ^2}}/{2 g^2}`
  `= 1/2 ({3 u sin theta}/{g} ± {u sqrt{9 sin^2 theta -8}}/{g})`
  `= {u}/{2g} (3 sin theta ± sqrt{9 sin^2 theta – 8})`

 

`text{S} text{ince} \ sqrt{9 sin^2 theta – 8} < sqrt{9 sin^2 theta} < 3 sin theta`

`=> \ text{in both solutions} \ \ t > 0`
 

`text{Consider the longer time:}`

`{u}/{2g} (3 sin theta + sqrt{9 sin^2 theta – 8})`

`< {u}/{2g} (3 sin  theta + sqrt{9 sin^2 theta – 8 sin^2 theta})`

`< {u}/{2g} (3 sin theta + sin theta)`

`< {2 u sin theta}/{g}`
 

`:. \ text{Longer time < time of flight}`

`:. \ text{Both times occur during time of flight.}`

Mechanics, EXT2 M1 2020 HSC 12b

A particle is projected from the origin with initial velocity `u` m/s at an angle  `theta`  to the horizontal. The particle lands at  `x = R`  on the `x`-axis. The acceleration vector is given by  `underset~a = ((0),(-g))`, where `g` is the acceleration due to gravity. (Do NOT prove this.)
 

  1. Show that the position vector  `underset~r (t)`  of the particle is given by
     
         `underset~r (t) = ((ut  cos theta),(ut  sin theta-frac{1}{2} g t^2))`.   (3 marks)

  2. Show that the Cartesian equation of the path of flights is given by
     
         `y = frac{-gx^2}{2u^2} (tan^2 theta-frac{2u^2}{gx} tan theta + 1)`.   (3 marks)

  3. Given  `u^2 > gR`, prove that there are 2 distinct values of  `theta`  for which the particle will land at  `x = R`.   (2 marks)

Show Answers Only
  1. `text{See Worked Solutions}`
  2. `text{See Worked Solutions}`
  3. `text{See Worked Solutions}`
Show Worked Solution

i.    `underset~a = ((0),(-g))`

`ddotx = 0`

`dotx = int ddotx \ dt = c`
 
`text{When} \ \ t = 0, \ dotx = u  cos theta \ =>  \ c = u  cos theta`

`=> dotx = u  cos theta`

 
`x = int dotx \ dt =u t cos theta + c`

`text{When} \ \ t = 0 , \ x = 0, \ c = 0`

`therefore \ x = ut  cos theta`
 

`ddoty = -g`

`doty = int-g \ dt = -g t + c`

`text{When} \ \ t = 0 , \ doty = u  sin theta`

`=> doty = u  sin theta-g t`
 

`y = int doty \ dt = ut  sin theta-frac {1}{2} g t^2 + c`

`text{When} \ \ t = 0, \ y = 0  \ => \ c = 0`

`therefore  y = ut  sin theta-frac(1)(2) g t^2`

`:. underset~r = ((x),(y)) = ((ut  cos theta),(ut sin theta-frac{1}{2} g t^2))`
 

ii.   `x = ut \ cos theta`

`t = frac{x}{u \ cos theta}`
 
`text{Substitute into} \ y:`

`y` `= u * frac{x}{u \ cos theta}\ sin theta-frac{1}{2} g ( frac{x}{u \ cos theta} )^2`
  `= x tan theta-frac{gx^2}{2 u^2 cos^2 theta}`
  `= frac{-gx^2}{2u^2} ( frac{1}{cos^2 theta}-frac{2u^2}{gx} tan theta )`
  `= frac{-gx^2}{2u^2} ( sec^2 theta-frac{2u^2}{gx} tan theta )`
  `= frac{-gx^2}{2u^2} ( tan^2 theta-frac{2u^2}{gx} tan theta + 1 )`

 

iii.    `text{When} \ \ x = R \ , \ y = 0`

Mean mark part (iii) 52%.
`frac{-gR^2}{2 u^2}` `( tan^2 theta-frac{2u^2}{gR} tan theta + 1 ) = 0`
   `tan^2 theta-frac{2u^2}{gR} tan theta + 1 = 0`
 
`Delta`		  
`= ( frac{-2u^2}{gR} )^2-4 * 1 * 1`
  `= frac{4u^4}{g^2 R^2}-4`

 

  `u^2` `> gR\ \ \ text{(given)}`
  `u^4` `> g^2 R^2`
  `frac{u^4}{g^2 R^2}` `> 1`
  `frac{4u^4}{g^2 R^2}` `> 4`
  `frac{4u^4}{g^2 R^2}-4` `> 0`
  `Delta` `> 0`

 
`therefore \ 2 \ text{distinct values of} \ \ theta\  \ text{satisfy} \ \ x = R.`

Mechanics, EXT2* M1 2019 HSC 13c

Two objects are projected from the same point on a horizontal surface. Object 1 is projected with an initial velocity of  `20\ text(ms)^(-1)` directed at an angle of  `pi/3`  to the horizontal. Object 2 is projected 2 seconds later.

The equations of motion of an object projected from the origin with initial velocity `v` at an angle `theta` to the `x`-axis are

`x = vt cos theta`

`y = -4.9t^2 + vt sin theta`,

where  `t`  is the time after the projection of the object. Do NOT prove these equations.

  1. Show that Object 1 will land at a distance  `(100 sqrt 3)/4.9` m from the point of projection.  (2 marks)

  2. The two objects hit the horizontal plane at the same place and time.

     

    Find the initial speed and the angle of projection of Object 2, giving your answer correct to 1 decimal place.  (3 marks)

Show Answers Only
  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `24.2\ text(ms)^(-1)`
Show Worked Solution

a.   `text(Object 1:)`

`x` `= 20t cos\ pi/3`
  `= 10t`
`y` `= -4.9t^2 + 20t sin\ pi/3`
  `= -4.9t^2 + 10 sqrt 3 t`

 
`text(Let)\ \ t_1 = text{time of flight (Object 1)}`

`-4.9t_1^2 + 10 sqrt 3 t_1` `= 0`
`t_1(-4.9t_1 + 10 sqrt 3)` `= 0`
`4.9t_1` `= 10 sqrt 3\ \ (t >= 0)`
`t_1` `= (10 sqrt 3)/4.9`

 
`text(Find)\ \ x\ \ text(when)\ \ t_1 = (10 sqrt 3)/4.9:`

`x` `= 10 xx (10 sqrt 3)/4.9`
  `= (100 sqrt 3)/4.9\ text(… as required)`

 

(ii)   `text{Time of flight (Object 2)}= (10 sqrt 3)/4.9 – 2`

♦ Mean mark 42%.

`text(Range)` `= (100 sqrt 3)/4.9`
`(100 sqrt 3)/4.9` `= v((10 sqrt 3)/4.9 – 2) cos theta`
`v cos theta` `= (100 sqrt 3)/4.9 xx 4.9/(10 sqrt 3 – 9.8)`
`v cos theta` `= (100 sqrt 3)/(10 sqrt 3 – 9.8) \ \ \ …\ (1)`

 

`0` `= -4.9t^2 + vt sin theta`
`0` `= -4.9 xx ((10 sqrt 3)/4.9 – 2)^2 + v((10 sqrt 3)/4.9 – 2) sin theta`
`0` `= -4.9((10 sqrt 3 – 9.8)/4.9) + v sin theta`
`v sin theta` `= 10 sqrt 3 – 9.8 \ \ \ …\ (2)`

 
`(2) ÷ (1)`

`tan theta` `= (10 sqrt 3 – 9.8) xx (10 sqrt 3 – 9.8)/(100 sqrt 3)`
  `= 0.3265…`
`:. theta` `= 18.1^@\ text{(1 d.p.)}`

 
`text{Substitute into (2)}`

`:.v` `= (10 sqrt 3 – 9.8) /(sin 18.1^@)`
  `= 24.206`
  `= 24.2\ text(ms)^(-1)\ text{(1 d.p.)}`

Mechanics, EXT2* M1 2017 HSC 13c

A golfer hits a golf ball with initial speed `V\ text(ms)^(−1)` at an angle `theta` to the horizontal. The golf ball is hit from one side of a lake and must have a horizontal range of 100 m or more to avoid landing in the lake.
 

     

Neglecting the effects of air resistance, the equations describing the motion of the ball are

`x = Vt costheta`

`y = Vt sintheta - 1/2 g t^2`,

where `t` is the time in seconds after the ball is hit and `g` is the acceleration due to gravity in `text(ms)^(−2)`. Do NOT prove these equations.

  1. Show that the horizontal range of the golf ball is
     
         `(V^2sin 2theta)/g` metres.  (2 marks)

  2. Show that if  `V^2 < 100 g`  then the horizontal range of the ball is less than 100 m.  (1 mark)

It is now given that  `V^2 = 200 g`  and that the horizontal range of the ball is 100 m or more.

  1. Show that  `pi/12 <= theta <= (5pi)/12`.  (2 marks)

  2. Find the greatest height the ball can achieve.  (2 marks)

Show Answers Only
  1. `text(See Worked Solution)`
  2. `text(See Worked Solution)`
  3. `text(See Worked Solution)`
  4. `25(2 – sqrt 3)\ text(metres)`
Show Worked Solution

i.   `text(Find)\ \ t\ \ text(when)\ \ y = 0:`

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

 

`text(Horizontal range)\ (x)\ text(when)\ \ t = (2Vsintheta)/g :`

`x` `= V · (2Vsintheta)/g costheta`
  `= (V^2 2sintheta costheta)/g`
  `= (V^2sin2theta)/g\ … text(as required)`

 

ii.   `text(If)\ \ V^2 < 100 g`

♦ Mean mark 44%.
`x` `< (100 g sin2theta)/g`
`x` `< 100 sin2theta`

 

`text(S)text(ince)\ −1 <= 2theta <= 1,`

`x < 100\ text(metres)`

 

iii.   `V^2 = 200g,\ \ x >= 100`

`(200 g · sin2theta)/g` `>= 100`
`sin2theta` `>= 1/2`

`:. pi/6 <= 2theta <= (5pi)/6`

`:. pi/12 <= theta <= (5pi)/12\ …\ text(as required)`

 

iv.   `text(Max height occurs when)`

♦♦ Mean mark 35%.
`t` `= 1/2 xx text(time of flight)`
  `= (Vsintheta)/g`

 
`text(Find)\ \ y\ \ text(when)\ \ t = (Vsintheta)/g`

`y` `= V · (Vsintheta)/g · sintheta – 1/2 g ((Vsintheta)/g)^2`
  `= (V^2 sin^2theta)/g – 1/2 · (V^2 sin^2 theta)/g`
  `= (V^2 sin^2theta)/(2g)`

 

`text(Max height when)\ theta = (5pi)/12\ (text(steepest angle)), V^2 = 200 g\ (text(given))`

`y_text(max)` `= (200 g · sin^2 ((5pi)/12))/(2g)`
  `= 100 sin^2 ((5pi)/12)`
  `= 50(1 – cos((5pi)/6))`
  `= 50(1 + sqrt3/2)`
  `= 25(2 + sqrt 3)\ text(metres)`

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^@`  

Mechanics, EXT2* M1 2014 HSC 14a

The take-off point `O` on a ski jump is located at the top of a downslope. The angle between the downslope and the horizontal is  `pi/4`.  A skier takes off from `O` with velocity `V` m s−1 at an angle `theta` to the horizontal, where  `0 <= theta < pi/2`.  The skier lands on the downslope at some point `P`, a distance `D` metres from `O`.
 

2014 14a
 

The flight path of the skier is given by

`x = Vtcos theta,\ y = -1/2 g t^2 + Vt sin theta`,      (Do NOT prove this.)

where  `t`  is the time in seconds after take-off.

  1. Show that the cartesian equation of the flight path of the skier is given by
     
         `y =  x tan theta - (gx^2)/(2V^2) sec^2 theta`.   (2 marks)

  2. Show that  
     
         `D = 2 sqrt 2 (V^2)/(g) cos theta (cos theta + sin theta)`.   (3 marks)

  3. Show that  
     
         `(dD)/(d theta) = 2 sqrt 2 (V^2)/(g) (cos 2 theta - sin 2 theta)`.   (2 marks)

  4. Show that `D` has a maximum value and find the value of `theta` for which this occurs.   (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

i.  `text(Show)\ \ y = x tan theta – (gx^2)/(2V^2) sec^2 theta`

`x` `= Vt cos theta`
`t` `= x/(V cos theta)\ \ \ …\ text{(1)}`

`text(Subst)\ text{(1)}\ text(into)\ y = -1/2 g t^2 + Vt sin theta`

`y` `= -1/2 g (x/(Vcostheta))^2 + V sin theta (x/(Vcostheta))`
  `= (-gx^2)/(2V^2 cos^2 theta) + x * (sin theta)/(cos theta)`
  `= x tan theta – (gx^2)/(2V^2) sec^2 theta\ \ \ text(… as required.)`

 

ii.  `text(Show)\ D = 2 sqrt 2 (V^2)/g\ cos theta (cos theta + sin theta)`

♦♦ Mean mark 28%
MARKER’S COMMENT: The key to finding `P` and solving for `D` is to realise you need the intersection of the Cartesian equation in (i) and the line `y=–x`.

`text(S)text(ince)\ P\ text(lies on line)\ y = -x`

`-x` `=x tan theta – (gx^2)/(2V^2) sec^2 theta`
`-1` `=tan theta – (gx)/(2V^2) sec^2 theta`
`(gx)/(2V^2) sec^2 theta` `= tan theta + 1`
`x (g/(2V^2))` `=(sin theta)/(cos theta) * cos^2 theta + 1 * cos^2 theta`
`x` `=(2V^2)/g\ (sin theta cos theta + cos^2 theta)`
  `=(2V^2)/g\ cos theta (cos theta + sin theta)`

 

`text(Given that)\ \ cos(pi/4)` `= x/D = 1/sqrt2`
`text(i.e.)\ \ D` `= sqrt 2 x`

 
`:.\ D = 2 sqrt 2 (V^2)/g\ cos theta (cos theta + sin theta)`

`text(… as required.)`

 

iii.  `text(Show)\ (dD)/(d theta) = 2 sqrt 2 (V^2)/g\ (cos 2 theta – sin 2 theta)`

`D` `= 2 sqrt 2 (V^2)/g\ (cos^2 theta + cos theta sin theta)`
`(dD)/(d theta)` `= 2 sqrt 2 (V^2)/g\ [2cos theta (–sin theta) + cos theta cos theta + (– sin theta) sin theta]`
  `= 2 sqrt 2 (V^2)/(g) [(cos^2 theta – sin^2 theta) – 2 sin theta cos theta]`
  `= 2 sqrt 2 (V^2)/g\ (cos 2 theta – sin 2 theta)\ \ \ text(… as required)`

 

iv.  `text(Max/min when)\ (dD)/(d theta) = 0`

♦ Mean mark 47%
IMPORTANT: Note that students can attempt every part of this question, even if they couldn’t successfully prove earlier parts.
`2 sqrt 2 (V^2)/g\ (cos 2 theta – sin 2 theta)` `= 0`
`cos 2 theta – sin 2 theta` `= 0`
`sin 2 theta` `= cos 2 theta`
`tan 2 theta` `= 1`
`2 theta` `= pi/4`
`theta` `= pi/8`
`(d^2D)/(d theta^2)` `= 2 sqrt 2 (V^2)/g\ [-2 sin 2theta – 2 cos 2 theta]`
  `= 4 sqrt 2 (V^2)/g\ (-sin 2 theta – cos 2 theta)`

 

`text(When)\ \ theta = pi/8:`

`(d^2 D)/(d theta^2)` `= 4 sqrt 2 (V^2)/g\ (- sin (pi/4) – cos (pi/4))`
  `= 4 sqrt 2 (V^2)/g\ (- 1/sqrt2 – 1/sqrt2) < 0`
  ` =>\ text(MAX)`

 

`:.\ D\ text(has a maximum value when)\ theta = pi/8`

Mechanics, EXT2* M1 2012 HSC 14b

A firework is fired from `O`, on level ground, with velocity `70` metres per second at an angle of inclination  `theta`. The equations of motion of the firework are

 `x = 70t cos theta\ \ \ \ `and`\ \ \ y = 70t sin theta\ – 4.9t^2`. (Do NOT prove this.) 

The firework explodes when it reaches its maximum height.
 

2012 14b
 

  1. Show that the firework explodes at a height of  `250 sin^2 theta`  metres.   (2 marks)

  2. Show that the firework explodes at a horizontal distance of  `250 sin 2 theta`  metres from  `O`.    (1 mark)

  3. For best viewing, the firework must explode at a horizontal distance between 125 m and 180 m from `O`, and at least 150 m above the ground.

     

    For what values of  `theta`  will this occur?   (3 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `66.97^@… <= theta <= 75^@\ text(for best viewing)`
Show Worked Solution

i.   `text(Show max height is)\ 250 sin^2 theta\ text(metres)`

`text(Firework explodes at max height)`

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

`y = 70t sin theta\ – 4.9t^2`

`dot y = 70 sin theta\ – 9.8t`
 

`text(When)\ \ dot y = 0`

`70 sin theta\ – 9.8t` `= 0`
`9.8t` `=70 sin theta`
`t` `= 70/9.8 sin theta`

 

`:.\ y_max` `= 70 * 70/9.8 sin theta * sin theta\ – 4.9 * (70^2)/(9.8^2) sin^2 theta`
  `= 500 sin^2 theta\ – 250 sin^2 theta`
  `= 250 sin^2 theta\ text(metres)\ \ \ \ \ text(… as required)`

 

ii.  `x = 70t cos theta`

`text(At)\ \ t = 70/9.8 sin theta`

`x` `= 70 * 70/9.8 sin theta cos theta`
  `= 250 * 2 sin theta cos theta`
  `= 250 sin 2 theta\ text(metres)\ \ \ \ \ text(… as required)`

 

iii.  `text(Best viewing when)`

♦♦ Mean mark 28%
MARKER’S COMMENT: Many students ignored the findings in earlier parts and tried unsuccessfully to solve inequalities that still contained `t`.

`125 <= x <= 180\ \ text(and)\ \ y >= 150`

`text(S)text(ince)\ x = 250 sin 2 theta`

`125` `<= 250 sin 2 theta <= 180`
`1/2` `<= sin 2 theta <= 18/25`

 

`text(In the 1st quadrant)`

`30^@ <= 2 theta <= 46.054…^@`

`15^@ <= theta <= 23.0^@`

`text(In the 2nd quadrant)`

`133.94…^@` `<= 2 theta <= 150^@`
`67.0^@` `<= theta <= 75^@`

 
`text{Using part (i)}`

`text(When)\ theta = 23.0^@`

`y_(max)` `= 250 xx sin^2 23.0^@`
  `~~ 38 < 150text(m)`

 
`=> text(“Highest” max height for)\ \ 15°<=theta<=23.0\ \ text(does not satisfy.)`

 
`text(When)\ theta = 67.0…^@`

`y_text(max)` `= 250 xx sin^2 67.0^@`
  `~~ 212 > 150 text(m)`

 

`=> text(“Lowest” max height for)\ \ 67°<=theta<=75°\ \ text(satisfies.)`

`:.\ 67.0^@ <= theta <= 75^@\ text( for best viewing)`.

Mechanics, EXT2* M1 2009 HSC 6a

Two points, `A` and `B`,  are on cliff tops on either side of a deep valley. Let `h` and `R` be the vertical and horizontal distances between `A` and `B` as shown in the diagram. The angle of elevation of `B` from `A` is  `theta`,  so that  `theta=tan^-1(h/R)`.
 

2009 6a
 

At time `t=0`,  projectiles are fired simultaneously from `A` and `B`.  The projectile from `A` is aimed at `B`, and has initial speed `U` at an angle of  `theta`  above the horizontal. The projectile from `B` is aimed at `A` and has initial speed `V` at an angle  `theta`  below the horizontal.

The equations of motion for the projectile from `A` are

`x_1=Utcos theta`   and   `y_1=Utsin theta-1/2 g t^2`,

and the equations for the motion of the projectile from `B` are

`x_2=R-Vtcos theta`   and   `y_2=h-Vtsin theta-1/2 g t^2`,     (DO NOT prove these equations.)

  1. Let `T` be the time at which  `x_1=x_2`.
     
    Show that  `T=R/((U+V)\ cos theta)`.   (1 mark)

  2. Show that the projectiles collide.    (2 marks)

  3. If the projectiles collide on the line  `x=lambdaR`,  where  `0<lambda<1`,  show that
     
         `V=(1/lambda-1)U`.    (1 mark)

Show Answers Only
  1. `text{Proof  (See Worked Solutions)}` 
  2. `text{Proof  (See Worked Solutions)}`
  3. `text{Proof  (See Worked Solutions)}`
Show Worked Solution

i.   `text(Let)\ \ x_1=x_2\ \ text(at)\ \ t=T`

`UTcos theta` `=R-VTcos theta`
`UTcos theta+VTcos theta` `=R`
`Tcos theta\ (U+V)` `=R`

 
 `:. T=R/((U+V)\ cos theta)\ \ text(… as required)`

 

ii.   `text(If particles collide,)\ \  y_1=y_2\ \ text(at)\ \ t=T`

♦♦♦ Mean mark data not available although “few” students were able to complete this part.
MARKER’S COMMENT: Better responses showed  `y_1=y_2`,  or  `y_1-y_2=0`  which eliminated  `–½ g t^2`  from the algebra. Substituting `tan theta=h//R` was a key element in completing this proof.
`y_1` `=UTsin theta-1/2 g T^2`
  `=(UR sin theta)/((U+V)\ cos theta)-1/2 g T^2`
`y_2` `=h-VTsin theta-1/2 g T^2`
  `=Rtan theta-(VR sin theta)/((U+V)\ cos theta)-1/2 g T^2`
  `=R(((sin theta/cos theta)(U+V) cos theta-V sin theta)/((U+V)\ cos theta))-1/2 g T^2`
  `=R/((U+V)\ cos theta)(Usin theta+Vsin theta-Vsin theta)-1/2 g T^2`
  `=(UR sin theta)/((U+V)\ cos theta)-1/2 g T^2`
  `=y_1`

 

`:.\ text(Particles collide since)\  y_1=y_2\ text(at)\ t=T`.

 

iii.  `text(Particles collide on line)\ \ x=lambdaR,\ text(i.e.  where)\ \ t=T`

MARKER’S COMMENT: Students must show sufficient working in proofs to convince markers they have derived the answer themselves.
`lambdaR` `=UTcos theta`
  `=U R/((U+V)\ cos theta)cos theta`
  `=(UR)/((U+V))`
`(U+V)(lambdaR)` `=UR`
`U+V` `=(UR)/(lambdaR)`
`V` `=U/lambda-U`
  `=(1/lambda-1)U\ \ text(… as required)`

Mechanics, EXT2* M1 2010 HSC 6b

A basketball player throws a ball with an initial velocity  `v`  m/s at an angle of  `theta`  to the horizontal. At the time the ball is released its centre is at  `(0,0)`, and the player is aiming for the point `(d,h)`  as shown on the diagram. The line joining  `(0,0) `  and  `(d,h)`  makes an angle  `alpha`  with the horizontal, where  `0<alpha<pi/2`.
 

6b
 

Assume that at time  `t`  seconds after the ball is thrown its centre is at the point  `(x,y)`,  where

`x=vtcos theta`

`y=vt sin theta-5 t^2`.      (DO NOT prove this.)

  1. If the centre of the ball passes through  `(d,h)`  show that
     
         `v^2=(5d)/(cos theta sin theta-cos^2 theta tan alpha)`    (3 marks)

    (2) What happens to  `v`  as  `theta\ ->pi/2` ?    (1 mark)

  2. For a fixed value of  `alpha`,  let  `F(theta)=cos theta sin theta-cos^2 theta tan alpha`.
     
    Show that  `F prime(theta)=0`   when   `tan2theta tan alpha=-1`  (2 marks)

  3. Using part (a)(ii)* or otherwise show that  `F prime(theta)=0`,   when  `theta=alpha/2+pi/4`.    (1 mark)

     

    *Please note for the purposes of this question, part (a)(ii) showed that  when  `tanA tanB=-1`,  then  `A-B=pi/2` 

  4. Explain why  `v^2`  is a minimum when  `theta=alpha/2+pi/4`     (2 marks)

Show Answers Only
  1. `text{Proof  (See Worked Solutions)}` 
  2. (1)  `v->oo`

     

    (2)  `v->oo`

  3. `text{Proof  (See Worked Solutions)}`
  4. `text{Proof  (See Worked Solutions)}` 
  5. `text{Explanation  (See Worked Solutions)}`
Show Worked Solution

i.   `text(Ball passes through)\ \ (d,h).`

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

♦ Mean mark 43%
IMPORTANT: It is critical to use the relationship `tan alpha=h/d` to achieve the required proof.
`d` `=vtcos theta`
`:. t` `=d/(vcos theta)`

 

`text(S)text(ince)\ \ y=h\ \ text(when)\ \ t=d/(vcos theta)`

`h` `=vtsin theta-5t^2`
  `=v(d/(vcos theta))sin theta-5(d^2)/(v^2cos^2 theta)`
`(5d^2)/(v^2cos^2 theta)` `=(dsin theta)/cos theta-h`
`(5d^2)/v^2` `=dsin theta cos theta-cos^2 thetaxxh`
`v^2/(5d^2)` `=1/(dsin theta cos theta-cos^2 thetaxxh)`
`v^2` `=(5d^2)/(dsin theta cos theta-cos^2 thetaxxh)`
  `=(5d)/(cos theta sin theta-cos^2 thetaxxh/d)`
  `=(5d)/(cos theta sin theta-cos^2 theta tan alpha)\ \ text(… as required)`

 

 

ii. (1)  `text(As)\ \  theta->alpha`

♦ Mean marks of 37% and 32% for parts (ii)(1) and (ii)(2) respectively.

`costheta sin theta-cos^2 theta tan alpha->0`

`v->oo`

ii. (2)  `text(As)\ \  theta->pi/2`

`cos theta sin theta-cos^2 theta tan theta->0`

`v->oo`

 

iii.  `text(Show)\ \ tan 2thetatan alpha=–1\ \ text(when)\ \ F prime(theta)=0`.

`F(theta)` `=costheta sin theta-cos^2theta tan alpha`
`F prime(theta)` `=cos theta costheta+sin theta (–sin theta)-2cos theta (–sin theta) tan alpha`
  `=cos^2 theta-sin^2 theta+sin 2theta tan alpha`
  `=cos 2theta+sin 2theta tan alpha`

 

`text(When)\ F prime(theta)=0`

♦♦ Mean mark part (iii) 21%
MARKER’S COMMENT: Many students failed to treat  `tan alpha` as a constant when differentiating.
`cos 2theta+sin2thetatan alpha` `=0`
`sin2thetatan alpha` `=-cos2theta`
`tan 2theta tan alpha` `=-1\ \ text(… as required)`

 

iv.  `text(Show that)\ \ F prime(theta)=0,\ \ text(when)\ \ theta=alpha/2+pi/4`

`text(If)\ \ tanA tanB=–1\ => A-B=pi/2`

`text{(Given in part (a)(ii) – see note in question)}`

 

`text(S)text(ince)\ \ tan2theta tan alpha=–1\ \ text{(see part (iii))}`

♦♦♦ Mean mark 14%
MARKER’S COMMENT: Linking part (a)(ii) to this solution was a feature of the more efficient and successful approaches.
`=>\ \ 2theta-alpha` `=pi/2`
`2theta` `=alpha+pi/2`
`theta` `=alpha/2+pi/4`

 

 `text(Given that)\ \ Fprime(theta)=0\ \ text(when)\ \ tan 2thetatan alpha=–1`

`:. Fprime(theta)=0\ \ text(when)\ \ theta=alpha/2+pi/4\ \ text(… as required)`

 

v.   `text(S)text(ince)\ \ v^2=(5d)/(F(theta)),`

`=>\ v^2\ \ text(is a MIN when)\ \  F(theta)\ \ text(is a MAX)`

`text(We know)\ \ F prime(theta)=0\ \ text(when)\ \ theta=alpha/2+pi/4`

♦♦♦ Mean mark 9%
MARKER’S COMMENT: Only a small minority of students  explained correctly that  `v^2`  is a MIN when  `F(theta)`  is a MAX.

`:.\ text(MAX or MIN when)\ \ theta=alpha/2+pi/4`

 

`F prime(theta)=cos 2theta+sin 2theta tan alpha`

`F ″(theta)=-2sin2theta+2cos2theta tan alpha`

`text(When)\ \ theta=alpha/2+pi/4`

`F″(alpha/2+pi/4)=-2sin(alpha+pi/2)+2cos(alpha+pi/2)tan alpha`

`text(S)text(ince)\ \ 0<alpha<pi/2`

`=> tan alpha>0`

`=>sin(alpha+pi/2)>0`

`=>cos(alpha+pi/2)<0`

`:. F″(alpha/2+pi/4)<0\ \ text(i.e.  MAX)`

`:.\ v^2\ \ text(is a minimum.)`