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STEP II 2007 question 11 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 2007 question 11 solution

i) \alpha = \tan^{-1}\frac{1}{2}, \; \theta = 60^{o}

\textbf{v} = -25\cos\alpha\cos\theta\textbf{i} + 25\cos\alpha\sin\theta\textbf{j} + (25\sin\alpha - 10t)\textbf{k}

\textbf{r} = (50 - 25\cos\alpha\cos\theta t)\textbf{i} + 25\cos\alpha\sin\theta t \textbf{j} + t(25\sin\alpha - 5 t)\textbf{k}

Finding the distance with Pythagoras' Theorem and then applying the identity \sin^{2}x + \cos^{2}x = 1 twice, we get:

D^{2} = 25\left(t^{4} - 10\sin\alpha t^{3} + 25t^{2} - 100\cos\alpha\cos\theta t + 100\right)

\cos\theta = \frac{1}{2}, \; \cos\alpha = \frac{2\sqrt{5}}{5}, \; \sin\alpha = \frac{\sqrt{5}}{5}

D^{2} = 25\left(t^{4} - 2\sqrt{5} t^{3} + 25t^{2} - 20\sqrt{5}t + 100\right) = 25(t^{2} - \sqrt{5}t + 10)^{2}

\Rightarrow D = 5(t^{2} - \sqrt{5}t + 10) = 5\left((t - \frac{\sqrt{5}}{2})^{2} + \frac{35}{4}\right)

Therefore the minimum distance from the origin is when t = \frac{\sqrt{5}}{2}

\textbf{P} = 37.5\textbf{i} + 12.5\sqrt{3}\textbf{j} + 6.25\textbf{k}

The horizontal bearing from O is \tan^{-1}\frac{12.5\sqrt{3}}{37.5} = 30^{o}.

ii) A particle is at its maximum height when the vertical component of its velocity is zero.

When t = \frac{\sqrt{5}}{2}, \textbf{v}_{k} = 0

iii) Using the position vector from part i), the distance between O and P is 43.75m. Since we are assuming no forces act on the bullet, it takes \frac{43.75}{350} = 0.125 seconds to reach P.

It is now necessary to create new velocity and position vectors for the particle with P as an origin.

\textbf{v} = -5\sqrt{5}\textbf{i} + 5\sqrt{15}\textbf{j} - 10t\textbf{k}

\textbf{r} = - 5\sqrt{5}t\textbf{i} + 5\sqrt{15}t\textbf{j} - 5t^{2}\textbf{k}

When t = 0.125, D^{2} = \frac{125}{64} + \frac{375}{64} + \frac{25}{64^{2}}

Ignoring the insignificant value contributed to the total by the change in height, we get:

D^{2} = \frac{500}{64}

Clearly this is between 7 and 8. It follows that D approximately equals 3m.

Solution by Dystopia.

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