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Two particles are fired simultaneously from the ground at time $t = 0$ - HSC - SSCE Mathematics Extension 1 - Question 6 - 2006 - Paper 1
Question 6
Two particles are fired simultaneously from the ground at time $t = 0$. Particle 1 is projected from the origin at an angle $\theta$, $0 < \theta < \frac{\pi}{2}$,... show full transcript
Worked Solution & Example Answer:Two particles are fired simultaneously from the ground at time $t = 0$ - HSC - SSCE Mathematics Extension 1 - Question 6 - 2006 - Paper 1
Step 1
Show that, while both particles are in flight, $L^2 = 2v^2t^2(1 - \sin \theta) - 2aV \cos \theta + a^2.$
Answer
To find the distance , we can express it in terms of the coordinates of both particles:
Squaring both sides gives us:
By expanding both terms and applying simplifications, we verify that we can arrive at:
$$L^2 = 2V^2t^2(1 - \sin \theta) - 2aV \cos \theta + a^2.$
Step 2
Show that the distance between the particles in flight is smallest when $t = \frac{a \cos \theta}{2V(1 - \sin \theta)}$.
Answer
To determine the time at which the distance is minimized, we need to differentiate with respect to and set the derivative equal to zero.
From the previously derived expression of , differentiate:
Solving for and simplifying gives:
Now substituting this back into results in the minimum distance:
Step 3
Show that the smallest distance between the two particles in flight occurs while Particle 1 is ascending if $V > \frac{aq \cos \theta}{\sqrt{2} \sin \theta(1 - \sin \theta)}.$
Answer
For Particle 1 to be ascending, its vertical velocity must be positive. The vertical velocity of Particle 1 is given by:
Plugging in the expression for derived earlier, we want:
Rearranging and simplifying leads to an inequality that confirms the condition for with respect to the other variables, thus validating the derived expression.