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Mechanics, EXT2* M1 2019 HSC 13d

The point  `O`  is on a sloping plane that forms an angle of 45° to the horizontal. A particle is projected from the point  `O`. The particle hits a point  `A`  on the sloping plane as shown in the diagram.
 


 

The equation of the line  `OA`  is  `y = -x`. The equations of motion of the particle are

`x = 18t`

`y = 18 sqrt(3t) - 5t^2,`

where  `t`  is the time in seconds after projection. Do NOT prove these equations.

  1. Find the distance  `OA`  between the point of projection and the point where the particle hits the sloping plane.  (2 marks)

  2. What is the size of the acute angle that the path of the particle makes with the sloping plane as the particle hits the point  `A`?  (3 marks)

Show Answers Only
  1. `(324(sqrt 2 + sqrt 6))/5\ text(units)`
  2. `30^@`
Show Worked Solution
i.    `x` `= 18t`
  `y` `= 18 sqrt 3 t – 5t^2`

 
`text(Particle hits slope when)\ \ y = -x`

`18 sqrt 3 t – 5t^2` `= -18t`
`5t^2 – 18t – 18 sqrt3 t` `= 0`
`t(5t – 18 – 18 sqrt 3)` `= 0`
`5t – 18 – 18 sqrt 3` `= 0`
`5t` `= 18 + 18 sqrt 3`
`t` `= (18 + 18 sqrt 3)/5`

 
`text(When)\ t = (18 + 18 sqrt 3)/5,`

`x = 18 xx ((18 + 18 sqrt 3)/5)`

`text{Using Pythagoras (isosceles Δ):}`

`OA` `= sqrt(2 xx 18^2 xx((18 + 18 sqrt 3)/5)^2)`
  `= sqrt 2 xx 18 xx ((18 + 18 sqrt 3)/5)`
  `= (324(sqrt 2 + sqrt 6))/5\ text(units)`

 

ii.    `x` `= 18t => dot x = 18`
  `y` `= 18 sqrt 3 t – 5t^2 => dot y = 18 sqrt 3 – 10t`

 
`text(When)\ \ t = (18 + 18 sqrt 3)/5,`

`dot y` `= 18 sqrt 3 – 10 ((18 + 18 sqrt 3)/5)`
  `= 18 sqrt 3 – 36 – 36 sqrt 3`
  `= -18 sqrt 3 – 36`
  `= -18(sqrt 3 + 2)`

 

`text(Find angle with the horizontal at impact:)`

♦ Mean mark part (ii) 39%.
 

 

`tan theta` `= (18(sqrt 3 + 2))/18`
  `= sqrt 3 + 2`
`theta` `= tan^(-1)(sqrt 3 + 2)`
  `= 75^@`

 
`:.\ text(Angle made with the slope)`

`= 75 – 45`

`= 30^@`

Mechanics, EXT2* M1 2005 6b

An experimental rocket is at a height of  5000 m, ascending with a velocity of  ` 200 sqrt 2\ text(m s)^-1`  at an angle of  45°  to the horizontal, when its engine stops.
 

 
After this time, the equations of motion of the rocket are:

`x = 200t`

`y = -4.9t^2 + 200t + 5000,`

where `t` is measured in seconds after the engine stops. (Do NOT show this.)

  1. What is the maximum height the rocket will reach, and when will it reach this height?  (2 marks)

  2. The pilot can only operate the ejection seat when the rocket is descending at an angle between 45° and 60° to the horizontal. What are the earliest and latest times that the pilot can operate the ejection seat?  (3 marks)

  3. For the parachute to open safely, the pilot must eject when the speed of the rocket is no more than  `350\ text(m s)^-1`. What is the latest time at which the pilot can eject safely?  (2 marks)

Show Answers Only
  1. `7041\ text(metres)`

     

    `20.4\ text(seconds.)`

  2. `40.8\ text(seconds)\ ;\ 55.8\ text(seconds)`
  3. `49.7\ text(seconds)`
Show Worked Solution

i.

    

`y = -4.9t^2 + 200t + 5000`

`dot y = -9.8t + 200`

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

`9.8t` `= 200`
`t` `= 20.408…`
  `= 20.4\ text{seconds  (to 1 d.p.)}`

 

`text(When)\ t = 20.408…`

`y` `= -4.9 (20.408…)^2 + 200 (20.408…) + 5000`
  `= 7040.816…`
  `= 7041\ text{m  (to nearest metre)}`

 

`:.\ text(The rocket will reach a maximum height of)`

`text(7041 metres when)\ \ t = 20.4\ text(seconds.)`

 

ii.  `text(The rocket is descending at 45° at point)\ A\ text(on the graph.)`

`text(The symmetry of the parabolic motion means that)\ \ A`

`text(occurs when)\ \ t = 2 xx text(time to reach max height, or)`

`40.8\ text(seconds.)`

 

`text(Point)\ B\ text(occurs when the rocket is descending at 60°)`

`text(to the horizontal.)`

 

`text(At)\ \ B,`

 

`-tan 60° = (dot y)/(dot x)`

`text{(The gradient of the projectile becomes}`

`text{negative after reaching its max height.)}`

`dot y = -9.8t + 200`

`dot x = d/(dt) (200t) = 200`

`:.\ – sqrt 3` `= (-9.8t + 200)/200`
`-200 sqrt 3` `= -9.8t + 200`
`9.8t` `= 200 + 200 sqrt 3`
`t` `= (200 + 200 sqrt 3)/9.8`
  `= (546.410…)/9.8`
  `= 55.756…`
  `= 55.8\ text{seconds  (nearest second)}`

 

`:.\ text(The pilot can operate the ejection seat)`

`text(between)\ \ t = 40.8 and t = 55.8\ text(seconds.)`

 

iii.

`v^2 = (dot x)^2 + (dot y)^2`

`text(When)\ \ v = 350`

`350^2` `= 200^2 + (-9.8t + 200)^2`
`122\ 500` `= 40\ 000 + (-9.8t + 200)^2`
`(-9.8t + 200)^2` `= 82\ 500`
`-9.8t + 200` `= +- sqrt (82\ 500)`
`9.8t` `= 200 +- sqrt (82\ 500)`
`t` `= (200 +- sqrt (82\ 500))/9.8`
  `= 49.717…\ \ \ (t > 0)`
  `= 49.7\ text{seconds  (to 1 d.p.)}`

 

`:.\ text(The latest time the pilot can eject safely)`

`text(is when)\ \ t = 49.7\ text(seconds.)`

Mechanics, EXT2* M1 2015 HSC 14a

A projectile is fired from the origin `O` with initial velocity `V` m s`\ ^(−1)` at an angle `theta` to the horizontal. The equations of motion are given by

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

Calculus in the Physical World, EXT1 2015 HSC 14a
 

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

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

  1. Find the angle that this projectile makes with the horizontal when
     
         `t = (2V)/(sqrt3\ g)`.  (2 marks)


  2. State whether this projectile is travelling upwards or downwards when
     
          `t = (2V)/(sqrt3\ g)`. Justify your answer.  (1 mark)

Show Answers Only
  1. `text(See Worked Solutions.)`
  2. `30^@`
  3. `text(Downwards)`
Show Worked Solution

i.   `text(Show range is)\ \ (V^2\ sin\ 2theta)/g`

`x` `= Vt\ cos\ theta`
`y` `= Vt\ sin\ theta − 1/2 g t^2`

 
`text(Horizontal range occurs 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`
`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\ sin\ 2theta)/g\ \ …\ text(as required)`

 

♦♦ Mean mark part (ii) 28%.
ii.    `x` `= Vt\ cos\ theta`
  `dot x` `= V\ cos\ theta`
  `y` `= Vt\ sin\ theta − 1/2 g t^2`
  `dot y` `= V\ sin\ theta − g t`

 

`text(When)\ \ t = (2V)/(sqrt3\ g)\ \ text(and)\ \ theta = pi/3`

`dot x` `= V\ cos\ pi/3 = V/2`
`dot y` `= V\ sin\ pi/3 − g\ (2V)/(sqrt3\ g)`
  `= (sqrt3V)/2 − (2V)/sqrt3`
  `= (sqrt3(sqrt3V) − 2 xx 2V)/(2sqrt3)`
  `= -V/(2sqrt3)`

Calculus in the Physical World, EXT1 2015 HSC 14a Answer

`text(Let)\ alpha =\ text(Angle of projectile with the horizontal)`

`tan\ α` `=(|\ doty\ |) / dotx`
  `= (V/(2sqrt3))/(V/2)`
  `= V/(2sqrt3) xx 2/V`
  `= 1/sqrt3`
`:.α` `= 30^@`

 
`:.\ text(When)\ \ t = (2V)/(sqrt3\ g),\ text(the projectile makes)`

`text(a)\ \ 30^@\ \ text(angle with the horizontal.)`

 

iii.  `text(When)\ t = (2V)/(sqrt3\ g)`

♦♦ Mean mark part (iii) 31%.

`dot y = −V/(2sqrt3)`

`text(The negative value of)\ \ dot y\ \ text(indicates that)`

`text(the particle is travelling downwards.)`

Mechanics, EXT2* M1 2011 HSC 6b

The diagram shows the trajectory of a ball thrown horizontally, at speed `v` m/s, from the top of a tower `h` metres above the ground level.
 

2011 6b
 

The ball strikes the ground at an angle of  45°, `d` metres from the base of the tower, as shown in the diagram. The equations describing the trajectory of the ball are

`x=vt`   and   `y=h-1/2 g t^2`,    (DO NOT prove this)

where `g` is the acceleration due to gravity, and `t` is time in seconds.

  1. Prove that the ball strikes the ground at time  
     
         `t=sqrt((2h)/(g))`  seconds.    (1 mark)

  2. Hence, or otherwise, show that  `d=2h`.    (2 marks)

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

i.   `text(Show)\ \ y=0\ \ text(when)\ t=sqrt((2h)/g)\ \ text(seconds:)`

`0` `=h-1/2 g t^2`
`1/2 g t^2` `=h`
`t^2` `=(2h)/g`
`:.t` `=sqrt((2h)/g)\ \ text(seconds,)\ \ t>=0\ \ text(… as required)`

 

ii.   `text(Show that)\ d=2h`

`x=vt\ \ \ \ =>\ \ \ dotx=v`

`y=h-1/2 g t^2\ \ \ \ =>\ \ \ doty=-g t`

`text(At)\ t=sqrt((2h)/g)`,

`doty` `=-gxxsqrt((2h)/g)`
  `=-sqrt(2gh)`

 

`text(S)text(ince the ball strikes the ground at)\ 45^@,\ text(we know)`

♦♦ Mean mark 33%
STRATEGY: The key to unlocking this proof is understanding that `tan theta`, where  `theta` is the angle of trajectory at impact of a projectile, equals  `|\ doty\ |/dotx`.
`tan45^@=` `|\ doty\ |/dotx`
`1=` `sqrt(2gh)/v`
`v=` `sqrt(2gh)`

 
`text(S)text(ince)\ \ x=d=vt\ \ text(when)\ \ t=sqrt((2h)/g)`

`d` `=sqrt(2gh)xxsqrt((2h)/g)`
  `=sqrt((2h)^2)`
  `=2h\ \ text( … as required)`