<s>[INST] Cambridge Admissions Exam: Where is the mistake? I am trying to solve the following problem.


My work:
Note: I am having only issues with (ii) but I have included my work for (i).
(i) I got the velocity  components as
$$ v_x=v\cos \theta \tag{Horizontal Velocity} $$
$$v_y=v\sin\theta - tg  \tag{Vertical Velocity}$$
We have $$ \tan \phi= \frac{v_y}{v_x}=\tan \theta -\frac{tg}{v \cos \theta}.$$
From the velocity components, we get the displacement components to be
$$s_x=vt\cos \theta \tag{Horizontal Displacement}$$
$$s_y=vt\sin \theta - t^2\frac{g}{2} \tag{Vertical Displacement}$$
By definition,
$$\tan \alpha = \frac{s_y}{s_x}=\tan \theta - \frac{tg}{2v\cos \theta} \tag{$*$}$$
Equating  the two expressions through $\frac{tg}{v\cos\theta}$ gives
$$\tan \theta + \tan \phi = 2\tan \alpha$$
as desired.
(ii) Here is where I start to have issues. From $(*)$ we have
$$\tan \alpha - \tan \theta = - \pm \frac{tg}{2v} \sqrt{1+\tan^2 \theta}$$
$$\implies 4\tan^2 \alpha + 4\tan^2 \theta-8\tan\theta \tan \alpha= \frac{t^2g^2}{v^2}(1+\tan^2\theta)$$
$$\iff (4-\frac{t^2g^2}{v^2})\tan^2\theta -8\tan \theta \tan \alpha +4\tan^2 \alpha -\frac{t^2g^2}{v^2}=0$$
If $\tan\theta$ and $\tan \theta'$ are the solutions for $\tan \theta$ in this equation, then
$$\tan \theta + \tan \theta'= \frac{8\tan \alpha}{4-\frac{t^2g^2}{v^2}}$$
$$ \tan \theta \tan \theta' = \frac{4\tan^2\alpha - \frac{t^2g^2}{v^2}}{4-\frac{t^2g^2}{v^2}}$$
and thus $$\tan(\theta + \theta')=\frac{\tan \theta + \tan \theta'}{1-\tan \theta \tan \theta'}=\frac{\frac{8\tan \alpha}{4-\frac{t^2g^2}{v^2}}}{1-\frac{4\tan^2\alpha - \frac{t^2g^2}{v^2}}{4-\frac{t^2g^2}{v^2}}}= \frac{\frac{8\tan \alpha}{4-\frac{t^2g^2}{v^2}}}{\frac{4-4\tan^2 \alpha}{4-\frac{t^2g^2}{v^2}}}=\tan{2\alpha}$$
which is wrong as we require $\tan(\theta + \theta')=-\cot \alpha$.
Does anybody know where my mistake is or why is this method wrong?
 [/INST] Please note that you are assuming that the time taken at both angles to reach $Q$ is same which is not correct assumption. We only know that the initial speed is same. So you should rather eliminate $t$ and write the projectile motion in terms of $x, y, u$ and $\theta$ where $x$ and $y$ are horizontal and vertical distance from $P$ to $Q$ respectively. $u$ is the initial speed.
$\displaystyle y = \tan\theta \cdot x - \frac{g}{2 \cdot u^2 \cdot \cos^2\theta} \cdot x^2 \ $
Rewriting in terms of $\tan \theta$,
$\displaystyle y = \tan\theta \cdot x - \frac{g \cdot (1+ \tan^2\theta)}{2 \cdot u^2} \cdot x^2$
$\displaystyle \frac{g \cdot x^2}{2 \cdot u^2} \cdot \tan^2\theta - x \cdot \tan \theta + (y + \frac{g \cdot x^2}{2 \cdot u^2}) = 0$
$\tan \theta + \tan \theta' = \displaystyle \frac{2 \cdot u^2}{g \cdot x}$
$\tan \theta \tan \theta' = \displaystyle 1 + \frac{2 \cdot u^2 \cdot y}{g \cdot x^2}$
Now using $ \ \tan (\theta+\theta') = \displaystyle \frac{\tan \theta + \tan \theta'} {1 - \tan \theta \tan \theta'}$, we get
$\tan (\theta + \theta') = \displaystyle - \frac{x}{y} = - \cot \alpha$
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