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2019 Paper 1 Q10
D: 1500.0 B: 1500.0
projectiles quadratic trigonometric cot sum of roots product of roots discriminant inequality greatest height optimisation

In this question, the \(x\)-axis is horizontal and the positive \(y\)-axis is vertically upwards. A particle is projected from the origin with speed \(u\) at an angle \(\alpha\) to the vertical. The particle passes through the fixed point \((h \tan \beta, h)\), where \(0 < \beta < 90^{\circ}\) and \(h > 0\).

  1. Show that $$c^2 - ck \cot \beta + 1 + k \cot^2 \beta = 0, \quad (*)$$ where \(c = \cot \alpha\) and \(k = \frac{2u^2}{gh}\). You are given that there are two distinct values of \(\alpha\) that satisfy equation \((*)\). Let \(\alpha_1\) and \(\alpha_2\) be these values.
    1. Show that $$\cot \alpha_1 + \cot \alpha_2 = k \cot \beta.$$ Show also that $$\alpha_1 + \alpha_2 = \beta.$$
    2. Show that $$k > 2(1 + \sec \beta).$$
  2. By considering the greatest height attained by the particle, show that \(k \geq 4 \sec^2 \alpha\).

Solution:

  1. The horizontal position of the particle at time \(t\) is \(u \sin\alpha t\), so \(T = \frac{h \tan \beta}{u \sin \alpha}\) The vertical position of the particle at this time \(T\) satisifes: \begin{align*} && h &= u \cos\alpha \frac{h \tan \beta}{u \sin\alpha} - \frac12 g \left ( \frac{h \tan \beta}{u \sin\alpha} \right)^2 \\ &&&= h\cot \alpha \tan \beta - \frac{gh^2}{2u^2} \tan^2 \beta \cosec^2 \beta \\ \Rightarrow && 1 &= c \tan \beta - \frac{1}{k} \tan^2 \beta (1 + c^2) \\ \Rightarrow && k \cot^2 \beta &= kc\cot \beta -1-c^2 \\ \Rightarrow && 0 &= c^2 -ck \cot \beta + 1 + k \cot^2 \beta \end{align*}
    1. As a quadratic in \(c\) the sum of the roots is \(k \cot \beta\), therefore \(\cot \alpha_1 + \cot \alpha_2 = k \cot \beta\). We also have that \(\cot \alpha_1 \cot \alpha_2 = 1 + k \cot^2 \beta\), so \begin{align*} && \cot (\alpha_1 + \alpha_2) &= \frac{\cot \alpha_1 \cot \alpha_2-1}{\cot \alpha_1 + \cot \alpha_2} \\ &&&= \frac{1 + k \cot^2 \beta - 1}{k \cot \beta} \\ &&&= \cot \beta \\ \Rightarrow && \beta &= \alpha_1 + \alpha_2 \pmod{\pi} \end{align*} but since \(\alpha_i \in (0, \frac{\pi}{2})\) the equation must hold exactly.
    2. Since it has two real roots we must have \begin{align*} && 0 &<\Delta = k^2 \cot^2 \beta - 4 (1 + k \cot^2 \beta) \\ &&&= k^2 \cot^2 \beta-4k \cot^2 \beta -4 \\ &&&= \cot^2 \beta (k^2 - 4k - 4(\sec^2 \beta - 1)) \\ &&&= \cot^2 \beta ( (k-2)^2 -4\sec^2 \beta) \\ \Rightarrow && k &> 2 + 2\sec \beta = 2(1+\sec \beta) \end{align*}
  2. The greatest height will satisfy \(v^2 = u^2 + 2as\) so \(0 = u^2 \cos^2 \alpha - 2gh_{max} \Rightarrow 4\sec^2 \alpha = \frac{2u^2}{gh_{max}} = k_{max}\), but this decreases with \(h\), so the smallest \(k\) can be is \(4\sec^2 \alpha\), ie \(k \geq 4 \sec^2 \alpha\)

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