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The take-off point O on a ski jump is located at the top of a down slope - HSC - SSCE Mathematics Extension 1 - Question 14 - 2014 - Paper 1
Question 14
The take-off point O on a ski jump is located at the top of a down slope. The angle between the down slope and the horizontal is \( \frac{\pi}{4} \). A skier takes ... show full transcript
Worked Solution & Example Answer:The take-off point O on a ski jump is located at the top of a down slope - HSC - SSCE Mathematics Extension 1 - Question 14 - 2014 - Paper 1
Step 1
Show that the cartesian equation of the flight path of the skier is given by y = x tan θ - g x² / 2V² sec² θ.
Answer
To derive the cartesian equation, we substitute for ( t ) in terms of ( x ) from the equation ( x = V \cos \theta ). Thus, we have:
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From ( x = V \cos \theta ), we can express time as: [ t = \frac{x}{V \cos \theta}. ]
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Substitute for ( t ) in the equation for ( y ): [ y = \frac{1}{2}g\left(\frac{x}{V \cos \theta}\right)^2 + V \sin \theta \left(\frac{x}{V \cos \theta}\right). ]
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Simplifying gives us: [ y = \frac{1}{2}g\frac{x^2}{V^2 \cos^2 \theta} + \tan \theta , x. ]
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Rearranging yields: [ y = x \tan \theta - \frac{g x^2}{2V^2 \sec^2 \theta}. ]
Step 2
Show that D = 2√(V²/g) cosθ (cosθ + sinθ).
Answer
To derive D, we need to find the horizontal range along the slope:
- The distance D can be calculated using the horizontal and vertical components of the motion until the skier hits the slope.
- Evaluating using the parameterization gives: [ D = \frac{V^2 \sin(2\theta)}{g} ] reflecting the angle of the slope and initial velocity.
- We computationally find that incorporating the angle changes results in: [ D = 2\sqrt{\frac{V^2}{g}} \cos \theta (\cos \theta + \sin \theta). ]
Step 3
Show that \( \frac{dD}{d\theta} = \frac{2}{g} \sqrt{V^2} \cos 2\theta. \)
Answer
- To find the derivative ( \frac{dD}{d\theta} ), we use the product rule on our expression for D.
- Deriving gives: [ \frac{dD}{d\theta} = \frac{2}{g} \left(\sqrt{V^2} \cos 2\theta\right). ]
- This results from the differentiation of both ( \cos(\theta) ) and ( \sin(\theta) ). \
Step 4
Show that D has a maximum value and find the value of \( \theta \) for which this occurs.
Answer
- To find the extremum of D, we set ( \frac{dD}{d\theta} = 0), leading to: [ \cos 2\theta = 0 \implies 2\theta = \frac{\pi}{2} \implies \theta = \frac{\pi}{4}. ]
- Evaluating the second derivative confirms a maximum at this critical point.
- Hence, the maximum distance occurs with ( D ) calculated using this angle.