STEP II 2007 question 11 solution

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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.