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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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1996 Paper 1 Q11
D: 1484.0 B: 1484.0
projectiles angle of projection speed of projection quadratic trigonometric parabolic trajectory algebraic manipulation

A particle is projected under the influence of gravity from a point \(O\) on a level plane in such a way that, when its horizontal distance from \(O\) is \(c\), its height is \(h\). It then lands on the plane at a distance \(c+d\) from \(O\). Show that the angle of projection \(\alpha\) satisfies \[ \tan\alpha=\frac{h(c+d)}{cd} \] and that the speed of projection \(v\) satisfies \[ v^{2}=\frac{g}{2}\left(\frac{cd}{h}+\frac{(c+d)^{2}h}{cd}\right)\,. \]

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1989 Paper 1 Q11
D: 1516.0 B: 1470.2
projectiles optimisation trigonometric range equation differentiation inequality angle of projection

A shot-putter projects a shot at an angle \(\theta\) above the horizontal, releasing it at height \(h\) above the level ground, with speed \(v\). Show that the distance \(R\) travelled horizontally by the shot from its point of release until it strikes the ground is given by \[ R=\frac{v^{2}}{2g}\sin2\theta\left(1+\sqrt{1+\frac{2hg}{v^{2}\sin^{2}\theta}}\right). \] The shot-putter's style is such that currently \(\theta=45^{\circ}\). Determine (with justification) whether a small decrease in \(\theta\) will increase \(R\). [Air resistance may be neglected.]

Solution: Notice that \(u_x = v \cos \theta, u_y = v \sin \theta\). We are interested in the time taken for the shot to hit the ground. \(-h = u_y t -\frac12 g t^2\) since our distance will be \(v \cos \theta \cdot t\). Solving this quadratic for \(t\) we obtain: \begin{align*} && 0 &= h + v \sin \theta \cdot t - \frac12 g \cdot t^2 \\ \Rightarrow && t_\pm &= \frac{-v \sin \theta \pm \sqrt{v^2 \sin^2 \theta+2hg}}{-g} \\ \Rightarrow && t_- &= \frac{v \sin \theta+v\sin \theta \sqrt{1 + \frac{2hg}{v^2 \sin^2 \theta}}}{g} \\ \Rightarrow && v \cos \theta t_{-} &= \frac{v^2}{g} \cos \theta \sin \theta \left( 1 + \sqrt{1 + \frac{2hg}{v^2 \sin^2 \theta}} \right) \\ &&&= \frac{v^2}{2g} \sin 2 \theta \left( 1 + \sqrt{1 + \frac{2hg}{v^2 \sin^2 \theta}} \right) \end{align*} Differentiating \(R\) wrt to \(\theta\) at \(\frac{\pi}{4}\) we obtain: \begin{align*} \frac{\d R}{\d \theta} &= \frac{v^2}{2g} \left (2 \cos 2 \theta + 2 \cos 2 \theta \sqrt{1 + \frac{2hg}{v^2 \sin^2 \theta}} + \sin 2\theta \left ( 1 + \frac{2hg}{v^2 \sin^2 \theta}\right)^{-\frac12} \frac12 \frac{2hg} {v^2}(-2) \frac{\cos \theta}{\sin^3 \theta}\right) \\ \frac{\d R}{\d \theta} \biggr \rvert_{\theta = \frac{\pi}{4}} &=\frac{v^2}{2g}\left(0+0- 4\left ( 1 + \frac{4hg}{v^2 }\right)^{-\frac12} \frac{hg} {v^2} \right) \\ &< 0 \end{align*} Therefore, since \(R\) is locally decreasing in \(\theta\) he should reduce his angle of projection slightly.

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