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Two points, A and B, are on cliff tops on either side of a deep valley - HSC - SSCE Mathematics Extension 1 - Question 6 - 2009 - Paper 1
Question 6
Two points, A and B, are on cliff tops on either side of a deep valley. Let h and R be the vertical and horizontal distances between A and B as shown in the diagram.... show full transcript
Worked Solution & Example Answer:Two points, A and B, are on cliff tops on either side of a deep valley - HSC - SSCE Mathematics Extension 1 - Question 6 - 2009 - Paper 1
Step 1
Let T be the time at which x1 = x2.
Answer
To show that T = ( \frac{R}{(U + V) \cos \theta} ), we start from the equations for the motion of the projectiles:
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From projectile A: [ x_1 = U T \cos \theta ]
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From projectile B: [ x_2 = R - V T \cos \theta ]
Setting these equal gives: [ U T \cos \theta = R - V T \cos \theta ]
Rearranging yields: [ U T \cos \theta + V T \cos \theta = R ]
Factoring out T gives: [ T \cos \theta (U + V) = R ]
Thus: [ T = \frac{R}{(U + V)\cos \theta} ]
Step 2
Show that the projectiles collide.
Answer
To show that the projectiles collide, we need to set their vertical positions equal at time T:
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The vertical position of projectile A at time T is: [ y_1 = U T \sin \theta - \frac{1}{2} g T^2 ]
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The vertical position of projectile B at time T is: [ y_2 = h - V T \sin \theta - \frac{1}{2} g T^2 ]
Setting these equal gives: [ U T \sin \theta - \frac{1}{2} g T^2 = h - V T \sin \theta - \frac{1}{2} g T^2 ]
Cancelling ( \frac{1}{2} g T^2 ) from both sides leads to: [ U T \sin \theta + V T \sin \theta = h ]
Thus: [ (U + V) T \sin \theta = h ]
From this, we can conclude the projectiles collide.
Step 3
If the projectiles collide on the line x = λR, where 0 < λ < 1, show that V = \frac{(1 - λ)}{λ} U.
Answer
Given that the projectiles collide on the line ( x = \lambda R ), we substitute ( x = \lambda R ) into the relevant position equations:
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From projectile A: [ \lambda R = U T \cos \theta \rightarrow T = \frac{\lambda R}{U \cos \theta} ]
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From projectile B: [ \lambda R = R - V T \cos \theta \rightarrow V T \cos \theta = R(1 - \lambda) ]
Now, solving for T gives: [ T = \frac{R(1 - \lambda)}{V \cos \theta} ]
Setting the two expressions for T equal: [ \frac{\lambda R}{U \cos \theta} = \frac{R(1 - \lambda)}{V \cos \theta} ]
Eliminating R and rearranging leads to: [ \frac{\lambda}{U} = \frac{(1 - \lambda)}{V} \Rightarrow V = \frac{(1 - \lambda)}{\lambda} U ]
Step 4
Sum the geometric series (1 + x^r) + (1 + x^r)y^1 + ... + (1 + x^r).
Answer
The sum of the geometric series can be approached as follows:
We take the series: [ S = (1 + x^r) + (1 + x^r)x + (1 + x^r)x^2 + ... + (1 + x^r)x^{n-1} ]
Factoring out ( (1 + x^r) ) gives: [ S = (1 + x^r)(1 + x + x^2 + ... + x^{n-1}) ]
Using the formula for the sum of a geometric series, we have: [ 1 + x + x^2 + ... + x^{n-1} = \frac{1 - x^n}{1 - x} ]
Thus, the overall sum becomes: [ S = (1 + x^r) \cdot \frac{1 - x^n}{1 - x} ]