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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 $ rac{ heta}{4}$. A skier takes off ... 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 - g x² / (2V² sec² θ)
Answer
To derive the required equation, we begin with the expressions given for the flight path:
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From the equations of motion, we have:
y = rac{1}{2} g t^2 + V ext{sin} heta imes tHere, is the time in seconds after take-off.
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Rearranging the first equation for , we get:
t = rac{x}{V ext{cos} heta}.
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Substituting this expression for into the second equation:
y = rac{1}{2} g igg( rac{x}{V ext{cos} heta} igg)^2 + V ext{sin} heta igg( rac{x}{V ext{cos} heta} igg).
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Simplifying this gives:
y = rac{g x^2}{2V^2 ext{cos}^2 heta} + x an heta.
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Finally, rearranging leads us to the expression:
y = x an heta - rac{g x^2}{2V^2 ext{sec}^2 heta}.
Step 2
Show that D = 2√2 V² / (g sin θ cos θ)
Answer
To derive the expression for D, we will utilize the results from part (i).
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Recognizing the equation for the flight path, we find where the skier lands on the slope given by:
D = x_{max} = rac{V^2 ext{sin}(2 heta)}{g}.
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The maximum distance D in terms of θ is obtained when is maximized, which occurs when , or .
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Thus, substituting gives:
D = 2 rac{V^2}{g}.
Step 3
Show that dD/dθ = 2√2 V² / (g sin 2θ)
Answer
To find the derivative of D with respect to θ, begin from the expression of D derived previously:
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Differentiating D with respect to θ using the chain rule:
rac{dD}{d heta} = rac{d}{d heta}igg( rac{V^2 ext{sin}(2 heta)}{g}igg).
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Applying the derivative gives:
rac{dD}{d heta} = rac{2V^2 ext{cos}(2 heta)}{g}.
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Recognizing that can be related as well leads to:
rac{dD}{d heta} = 2 rac{ ext{V}^2}{g ext{sin}(2 heta)}.
Step 4
Show that D has a maximum value and find the value of θ for which this occurs.
Answer
To find the maximum value of D, we set rac{dD}{d heta} = 0:
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From part (iii), set:
2 rac{V^2}{g ext{sin}(2 heta)} = 0.
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This indicates that , giving:
2 heta = n rac{ ext{π}}{2}, ext{ } n ext{ integer}.
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The maximum occurs at:
.
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However, as we seek a valid angle in the context of the problem, the value of θ for maximum distance occurs at 45°, resulting in peak D.