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 $\frac{\pi}{4}$. A skier takes off from O with velocity $V \text{ m s}^{-1}$ at an angle $\theta$ to the horizontal, where $0 \leq \theta \leq \frac{\pi}{2}$. The skier lands on the downslope at some point P, a distance D metres from O. The flight path of the skier is given by $x = V \cos \theta, \quad y = \frac{1}{2} g t^2 + V \sin \theta \cdot t$. (Do NOT prove this.) (i) Show that the cartesian equation of the flight path of the skier is given by $y = x \tan \theta - \frac{g x^2}{2V^2 \cos^2 \theta}.$ (ii) Show that $D = 2 \sqrt{\frac{V^2}{g}} \cos \theta (\cos \theta + \sin \theta).$ (iii) Show that $\frac{dD}{d\theta} = 2\sqrt{\frac{V^2}{g}} \cos 2\theta.$ (iv) Show that D has a maximum value and find the value of $\theta$ for which this occurs.

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The take-off point O on a ski jump is located at the top of a downslope - 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 downslope. The angle between the downslope and the horizontal is $\frac{\pi}{4}$. A skier takes off fro... show full transcript

Worked Solution & Example Answer:The take-off point O on a ski jump is located at the top of a downslope - 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 - rac{g x²}{2V² cos² θ}.

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Answer

To derive the cartesian equation, we start by substituting $t$ from the equation of $x$ . From the flight path, we have:

$x = V \cos \theta \cdot t \implies t = \frac{x}{V \cos \theta}.$

Substituting this into the equation for $y$ gives:

$y = \frac{1}{2} g \left(\frac{x}{V \cos \theta}\right)^2 + V \sin \theta \cdot \left(\frac{x}{V \cos \theta}\right).$

This simplifies to:

$y = \frac{g x^2}{2V^2 \cos^2 \theta} + x \tan \theta,$

which rearranges to:

$y = x \tan \theta - \frac{g x^2}{2V^2 \cos^2 \theta}.$

Step 2

Show that D = 2√2 V² / g cos θ (cos θ + sin θ).

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Answer

From the previous section, the distance $D$ can be calculated as the horizontal range. To find $D$ , we need:

$D = rac{V^2 \sin 2\theta}{g}.$

Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$ , we can rewrite this as:

$D = \frac{V^2 \cdot 2 \sin \theta \cos \theta}{g} = \frac{2V^2}{g} \cos \theta (\cos \theta + \sin \theta).$

Step 3

Show that \frac{dD}{dθ} = 2√{\frac{V²}{g}} cos 2θ.

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Answer

To find the derivative of D with respect to $\theta$ , we apply the product rule on the expression:

$D = \frac{2V^2}{g} \cos \theta (\sin \theta + \cos \theta).$

Differentiating gives:

$\frac{dD}{d\theta} = \frac{2V^2}{g} \left( -\sin \theta (\sin \theta + \cos \theta) + \cos \theta (\cos \theta) \right) = 2\sqrt{\frac{V^2}{g}} \cos 2\theta.$

Step 4

Show that D has a maximum value and find the value of θ for which this occurs.

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Answer

To find the maximum value of D, we set $\frac{dD}{d\theta} = 0$ :

$\cos 2\theta = 0.$

This occurs at:

$2\theta = \frac{\pi}{2} + n\pi \implies \theta = \frac{\pi}{4} + \frac{n\pi}{2}.$

As $0 \leq \theta \leq \frac{\pi}{2}$ , the critical angle for maximum D is $heta = \frac{\pi}{4}$ . Thus, D has a maximum value at this angle.

The take-off point O on a ski jump is located at the top of a downslope - HSC - SSCE Mathematics Extension 1 - Question 14 - 2014 - Paper 1

Worked Solution & Example Answer:The take-off point O on a ski jump is located at the top of a downslope - HSC - SSCE Mathematics Extension 1 - Question 14 - 2014 - Paper 1

Show that the cartesian equation of the flight path of the skier is given by y = x tan θ - rac{g x²}{2V² cos² θ}.

Show that D = 2√2 V² / g cos θ (cos θ + sin θ).

Show that \frac{dD}{dθ} = 2√{\frac{V²}{g}} cos 2θ.

Show that D has a maximum value and find the value of θ for which this occurs.

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