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David is in a life raft and Anna is in a cabin cruiser searching for him - HSC - SSCE Mathematics Extension 1 - Question 7 - 2003 - Paper 1
Question 7
David is in a life raft and Anna is in a cabin cruiser searching for him. They are in contact by mobile telephone. David tells Anna that he can see Mt Hope. From Dav... show full transcript
Worked Solution & Example Answer:David is in a life raft and Anna is in a cabin cruiser searching for him - HSC - SSCE Mathematics Extension 1 - Question 7 - 2003 - Paper 1
Step 1
Find the distance and bearing of the life raft from Anna's position.
Answer
To solve this part, we first need to determine the coordinates of points A (David's position) and H (top of Mt Hope).
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Coordinates of A:
- The bearing from Anna to Mt Hope is 139°. This corresponds to an angle of 139° measured clockwise from north.
- The angle of elevation to the top of the mountain, H, is 23° and the height is 1500 m. Using trigonometry, we can find the horizontal distance from Anna to the base of Mt Hope:
Let the horizontal distance from point A to point B (the base of Mt Hope) be x.
The relationship can be formed using:
an(23°) = rac{1500}{x} \{
Rearranging gives: x = rac{1500}{ an(23°)}.
- Coordinates of H:
- Using the angle of elevation, calculate:
.
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Determine the bearing from Anna’s position to David in the life raft:
- The bearing from point B to point D (David) is based on the angle of elevation from David’s perspective (16°) at a bearing of 109°.
- We calculate the final coordinates of David and find the distance (d) from Anna’s coordinates (A) to David’s coordinates (D).
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Final Results: Calculated distance d will be expressed in meters and the bearing will follow the proper convention to be presented safely.
Step 2
(i) Show that the maximum height reached, h metres, is given by h = v^2 sin^2 α / 2g.
Answer
To determine the maximum height reached by the particle.
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The y-component of the velocity can be expressed as: .
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According to the equations of motion, at the maximum height, the vertical component of velocity will be zero: .
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We rearrange for time t: t = rac{v imes ext{sin}( heta)}{g}.
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Substituting this time value into the vertical motion equation: y_{max} = v imes ext{sin}( heta) t - rac{1}{2} g t^2 gives the height as:
h = rac{v^2 ext{sin}^2( heta)}{2g}.
Step 3
(ii) Show that it returns to the initial height at x = v^2 / g sin 2α.
Answer
To show the horizontal distance covered when the projectile returns to the initial height.
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Use the formula for x:
x = v imes ext{cos}( heta) imes t \. -
Substitute for time using the vertical motion formula, focusing on returning to initial height. When y = 0, simplify the calculations with complete substitution to determine the expression in terms of total velocity and angles involved, arriving at:
x = rac{v^2 ext{sin}(2 heta)}{g}.
Step 4
(iii) Show that the maximum separation, d, that Chris and Sandy can have and still catch the ball is given by d = 4x(H − S) if v^2 ≥ 4g(H − S), and d = v^2 / g if v^2 < 4g(H − S).
Answer
To find the maximum separation:
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When the ball is thrown, it travels upward and then downward back to the height S. The maximum separation occurs during its travel.
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Calculate the times it takes to reach the maximum height and to come back:
- Time of flight to maximum height is given by t_1 = rac{v imes sin( heta)}{g}.
- Time to return to S is equal to initial velocity determination.
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Using the calculated distance relationships, showing effective motion and evaluating special cases depending on the initial velocity provides equations returning: when velocity conditions are fulfilled, else use a simplified direct relationship: d = rac{v^2}{g}.