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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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2014 Paper 2 Q10
D: 1600.0 B: 1484.0
projectiles trajectory equation envelope optimisation quadratic in parameter curve sketching range parabola

A particle is projected from a point \(O\) on horizontal ground with initial speed \(u\) and at an angle of \(\theta\) above the ground. The motion takes place in the \(x\)-\(y\) plane, where the \(x\)-axis is horizontal, the \(y\)-axis is vertical and the origin is \(O\). Obtain the Cartesian equation of the particle's trajectory in terms of \(u\), \(g\) and~\(\lambda\), where \(\lambda=\tan\theta\). Now consider the trajectories for different values of \(\theta\) with \(u\)~fixed. Show that for a given value of~\(x\), the coordinate~\(y\) can take all values up to a maximum value,~\(Y\), which you should determine as a function of \(x\), \(u\) and~\(g\). Sketch a graph of \(Y\) against \(x\) and indicate on your graph the set of points that can be reached by a particle projected from \(O\) with speed \(u\). Hence find the furthest distance from \(O\) that can be achieved by such a projectile.

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2012 Paper 2 Q9
D: 1600.0 B: 1500.0
projectiles inequality quadratic trigonometric trajectory equation range net clearance optimisation

A tennis ball is projected from a height of \(2h\) above horizontal ground with speed \(u\) and at an angle of \(\alpha\) below the horizontal. It travels in a plane perpendicular to a vertical net of height \(h\) which is a horizontal distance of \(a\) from the point of projection. Given that the ball passes over the net, show that \[ \frac 1{u^2}< \frac {2(h-a\tan\alpha)}{ga^2\sec^2\alpha}\,. \] The ball lands before it has travelled a horizontal distance of \(b\) from the point of projection. Show that \[ \sqrt{u^2\sin^2\alpha +4gh \ } < \frac{bg}{u\cos\alpha} + u \sin\alpha\,. \] Hence show that \[ \tan\alpha < \frac{h(b^2-2a^2)}{ab(b-a)}\,. \]

Solution: \begin{align*} && s &= ut \\ \Rightarrow && a &= u \cos \alpha t\\ \Rightarrow && t &= \frac{a}{u \cos \alpha}\\ && s &= ut+ \frac12at^2 \\ \Rightarrow && -h &< -u\sin \alpha \frac{a}{u \cos \alpha}-\frac12 g \left (\frac{a}{u \cos \alpha} \right)^2 \\ &&&= -a \tan \alpha-\frac12 g a^2 \frac{1}{u^2} \sec^2 \alpha \\ \Rightarrow && \frac12 g a^2 \frac{1}{u^2} \sec^2 \alpha &< h -a\tan \alpha \\ \Rightarrow &&\frac{1}{u^2} &< \frac{2(h-a\tan \alpha)}{ga^2 \sec^2 \alpha} \end{align*} \begin{align*} && s &= ut + \frac12a t^2 \\ \Rightarrow && 2h &= u\sin \alpha t + \frac12 gt^2 \\ \Rightarrow && t &= \frac{-u\sin \alpha \pm \sqrt{u^2 \sin^2 \alpha+4hg}}{g}\\ && t &= \frac{-u\sin \alpha +\sqrt{u^2 \sin^2 \alpha+4hg}}{g}\\ && s &= ut \\ \Rightarrow && b &> u \cos \alpha t \\ \Rightarrow && \frac{b}{u \cos \alpha} &> \frac{-u\sin \alpha +\sqrt{u^2 \sin^2 \alpha+4hg}}{g} \\ \Rightarrow && \sqrt{u^2 \sin^2 \alpha+4hg} &< \frac{bg}{u \cos \alpha} + u \sin \alpha \\ \end{align*} \begin{align*} \Rightarrow && u^2 \sin^2 \alpha+4hg &< \frac{b^2g^2}{u^2 \cos^2 \alpha} +u^2 \sin^2 \alpha + 2bg \tan \alpha \\ \Rightarrow && 4hg - 2bg \tan \alpha &< \frac{b^2g^2}{u^2 \cos^2 \alpha} \\ &&&< \frac{b^2g^2}{\cos^2 \alpha} \frac{2(h-a\tan \alpha)}{ga^2 \sec^2 \alpha} \\ &&&= \frac{2b^2g(h-a\tan \alpha)}{a^2} \\ \Rightarrow && \tan \alpha \left (\frac{2b^2g}{a} - 2bg \right) &< \frac{2b^2gh}{a^2} - 4hg \\ \Leftrightarrow && \tan \alpha \left (\frac{2b^2g- 2abg}{a} \right) &< \frac{2b^2gh- 4hga^2}{a^2} \\ \Leftrightarrow && \tan \alpha \left (\frac{2bg(b- a)}{a} \right) &< \frac{2hg(b^2- 2a^2)}{a^2} \\ \Rightarrow && \tan \alpha &< \frac{h(b^2-2a^2)}{ab(b-a)} \end{align*}

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2011 Paper 1 Q9
D: 1516.0 B: 1484.7
projectiles trajectory equation quadratic trigonometry tan θ range algebraic manipulation horizontal plane

A particle is projected at an angle \(\theta\) above the horizontal from a point on a horizontal plane. The particle just passes over two walls that are at horizontal distances \(d_1\) and \(d_2\) from the point of projection and are of heights \(d_2\) and \(d_1\), respectively. Show that \[ \tan\theta = \frac{d_1^2+d_\subone d_\subtwo +d_2^2}{d_\subone d_\subtwo}\,. \] Find (and simplify) an expression in terms of \(d_1\) and \(d_2\) only for the range of the particle.

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2003 Paper 1 Q9
D: 1500.0 B: 1500.0
projectiles quadratic in tan trajectory equation discriminant inequality sum of angles trigonometric identity optimisation

A particle is projected with speed \(V\) at an angle \(\theta\) above the horizontal. The particle passes through the point \(P\) which is a horizontal distance \(d\) and a vertical distance \(h\) from the point of projection. Show that \[ T^2 -2kT + \frac{2kh}{d}+1=0\;, \] where \(T=\tan\theta\) and \(\ds k= \frac{V^2}{gd}\,\). %Derive an equation relating \(\tan \theta\), \(V\), \(g\), \(d\) and \(h\). Show that, if \(\displaystyle {kd > h + \sqrt {h^2 + d^2}}\;\), there are two distinct possible angles of projection. Let these two angles be \(\alpha\) and \(\beta\). Show that \(\displaystyle \alpha + \beta = \pi - \arctan ( {d/ h}) \,\).

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