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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
Show that T = \frac{R}{(U + V) \cos \theta}
Answer
To show that T = \frac{R}{(U + V) \cos \theta}, we need to equate the horizontal distances traveled by each projectile.
For the projectile from A:
[ x_1 = U T \cos \theta ]
For the projectile from B:
[ x_2 = R - V T \cos \theta ]
Setting these two expressions equal gives:
[ U T \cos \theta = R - V T \cos \theta ]
Rearranging this:
[ UT \cos \theta + VT \cos \theta = R ]
Factoring out the T:
[ T (U + V) \cos \theta = 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 verify that at some time T, both projectiles have the same coordinates (x, y).
Setting the equations for y equal:
[ U T \sin \theta - \frac{1}{2} g T^2 = h - V T \sin \theta - \frac{1}{2} g T^2 ]
Simplifying yields:
[ U T \sin \theta + V T \sin \theta = h ]
Factoring T out:
[ T (U + V) \sin \theta = h ]
Thus, both projectiles are at the same vertical position when:
[ T = \frac{h}{(U + V) \sin \theta} ]
If we substitute T back into x equations, and find both equal x, the projectiles do collide.
Step 3
Show that V = \left( \frac{1}{\lambda} - 1 \right) U.
Answer
To show this relationship, we use the previous results where we set the x-coordinate of collision at x = \lambda R.
From the equation of projectile from A:
[ \lambda R = U T \cos \theta ]
And from B:
[ \lambda R = R - V T \cos \theta ]
We can rearrange the second to get:
[ V T \cos \theta = R(1 - \lambda) ]
Now replacing T from the previous part gives:
[ V \cos \theta \cdot \frac{R}{(U + V) \cos \theta} = R(1 - \lambda) ]
Thus,
[ V = \left( \frac{(U + V)(1 - \lambda)}{1} \right) ]
Giving:
[ V = \left( \frac{1}{\lambda} - 1 \right) U ]
This completes the proof.