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2017 Paper 1 Q9
D: 1484.0 B: 1484.0
projectiles optimisation trigonometric minimum speed angle of projection trajectory equation differentiation

A particle is projected at speed \(u\) from a point \(O\) on a horizontal plane. It passes through a fixed point \(P\) which is at a horizontal distance \(d\) from \(O\) and at a height \(d \tan \beta\) above the plane, where \(d>0\) and \(\beta \) is an acute angle. The angle of projection \(\alpha\) is chosen so that \(u\) is as small as possible.

  1. Show that \(u^2 = gd \tan \alpha\) and \(2\alpha = \beta + 90^\circ\,\).
  2. At what angle to the horizontal is the particle travelling when it passes through \(P\)? Express your answer in terms of \(\alpha\) in its simplest form.

Solution:

  1. \(\,\) \begin{align*} && d &= u \cos \alpha t \\ && d \tan \beta &= u \sin \alpha t - \frac12 gt^2 \\ && &= d\tan \alpha - \frac1{2u^2} g d^2 \sec^2 \alpha \\ \Rightarrow && u^2 &= \frac{gd \sec^2 \alpha}{2(\tan \alpha + \tan \beta)} \\ &&&= \frac{gd t^2}{2(t+\tan \beta)} \\ && \frac{\d}{\d t} \left (u^2 \right) &= \frac{2gdt\cdot 2(t+\tan \beta) - gdt^2 \cdot 2}{4(t+\tan \beta)^2} \\ &&&= \frac{2gdt(2t-t+2\tan \beta)}{4(t+\tan \beta)^2} \\ &&&= \frac{gdt(t+2\tan \beta)}{2(t+\tan \beta)^2} \\ \end{align*} So either \(t = 0\) or \(t = -2 \tan \beta\) \begin{align*} && u^2 &= \frac{gd\cdot 4 \tan^2 \beta}{2(-2\tan \beta + \tan \beta)} \\ &&&= \frac{2gd \tan \beta}{-1} \\ &&&= gd (-2\tan \beta) \\ &&&= gd \tan \alpha \end{align*} \begin{align*} && d \tan \beta &= d \tan \alpha - \frac12 \frac{gd^2}{gd \tan \alpha \cdot \cos^2 \alpha } \\ \Rightarrow && \tan \beta&= \tan \alpha - \frac{1}{2\sin \alpha \cos \alpha} \\ &&&= \frac{2 \sin^2 \alpha - 1}{2 \sin \alpha \cos \alpha} \\ &&&= \frac{-\cos 2 \alpha}{\sin 2 \alpha} \\ &&&= -\cot 2 \alpha \\ &&&= \tan (2\alpha - 90^\circ) \\ \Rightarrow && \beta &= 2\alpha - 90^\circ \\ \Rightarrow && 2\alpha &= \beta + 90^\circ \end{align*}
  2. Suppose the angle to the horizontal is \(\theta\), then \(\tan \theta = \frac{v_y}{v_x}\) so \begin{align*} && \tan \theta &= \frac{u \sin \alpha - gt}{u \cos \alpha} \\ &&&= \frac{u \sin \alpha - \frac12 g \frac{d}{u \cos \alpha}}{u \cos \alpha} \\ &&&= \frac{u^2\sin \alpha \cos \alpha - gd}{u^2 \cos^2 \alpha} \\ &&&= \frac{gd \tan \alpha \sin \alpha \cos \alpha- gd}{ gd \tan \alpha \cdot \cos^2 \alpha} \\ &&&= \frac{\tan \alpha \sin \alpha \cos \alpha - 1}{ \sin \alpha \cos \alpha} \\ &&&= \frac{\sin^2 \alpha - 1}{\sin \alpha \cos \alpha} \\ &&&= -\frac{\cos \alpha}{\sin \alpha}\\ &&&= - \cot \alpha = \tan (\alpha - 90^\circ)\\ \Rightarrow && \theta &= \alpha - 90^\circ \end{align*}

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