The diagram shows a shot put being projected with a velocity $v\ \mathrm{ms}^{-1}$ from the point $O$ at an angle $\theta$ made with the horizontal. The point $O$ is $1.5\ \mathrm{m}$ above the point $A$ on the ground. The $x$-$y$ plane is taken to be the plane that contains the trajectory of this projectile motion with $x$-axis parallel to the horizontal and $O$ being the origin. The equation of the trajectory of this projectile motion is known to be $y = x\tan\theta - \dfrac{g x^2}{2v^2\cos^2\theta},$ where $g\ \mathrm{ms}^{-2}$ is the acceleration due to gravity. The constant $g$ is taken to be 10 and the distance between $A$ and $B$ is denoted by $h\ \mathrm{m}$.

1. Identify the coordinates of the landing point and substitute them into the trajectory equation.

From the given context, the origin $O$ is $1.5\text{ m}$ above the ground point $A$. Thus, the ground lies at $y = -1.5$.
At point $B$, the horizontal distance from $A$ is $h$, giving its coordinates as $(h, -1.5)$.
Substitute $x = h$, $y = -1.5$, $v = 10$, and $g = 10$ into the trajectory equation $y = x\tan\theta - \dfrac{g x^2}{2v^2\cos^2\theta}$:
$-1.5 = h\tan\theta - \dfrac{10h^2}{2(10)^2\cos^2\theta}$

2. Simplify the equation by clearing fractions.

Express $\tan\theta$ as $\dfrac{\sin\theta}{\cos\theta}$ and simplify the fraction on the right:
$-1.5 = \dfrac{h\sin\theta}{\cos\theta} - \dfrac{10h^2}{200\cos^2\theta}$
$-1.5 = \dfrac{h\sin\theta}{\cos\theta} - \dfrac{h^2}{20\cos^2\theta}$
Multiply the entire equation by $20\cos^2\theta$ to eliminate the denominators:
$-30\cos^2\theta = 20h\sin\theta\cos\theta - h^2$
Rearranging the terms to bring $h^2$ to the front with a positive coefficient:
$h^2 - 20h\sin\theta\cos\theta - 30\cos^2\theta = 0$

3. Apply double angle identities to reach the final 'shown' form.

Recall the double angle identities $\sin 2\theta = 2\sin\theta\cos\theta$ and $\cos 2\theta = 2\cos^2\theta - 1$ (which rearranges to $2\cos^2\theta = \cos 2\theta + 1$).
Factor out constants to match these identities:
$h^2 - 10h(2\sin\theta\cos\theta) - 15(2\cos^2\theta) = 0$
Substitute the double angle identities into the equation:
$h^2 - 10h\sin 2\theta - 15(\cos 2\theta + 1) = 0$
Expand and simplify to obtain the required equation:
$h^2 - 10h\sin 2\theta - 15\cos 2\theta - 15 = 0 \quad \text{(shown)}$

1. Differentiate the equation from part (a) implicitly with respect to $\theta$.

Given the equation from (a):
$h^2 - 10h\sin 2\theta - 15\cos 2\theta - 15 = 0$

Differentiating both sides with respect to $\theta$:
$2h \frac{dh}{d\theta} - \left( 10 \frac{dh}{d\theta} \sin 2\theta + 10h(2\cos 2\theta) \right) - 15(-2\sin 2\theta) - 0 = 0$
$2h \frac{dh}{d\theta} - 10 \frac{dh}{d\theta} \sin 2\theta - 20h\cos 2\theta + 30\sin 2\theta = 0$

2. Apply the condition for a stationary value and express $h$ in terms of $\theta$.

For a stationary value of $h$ as $\theta$ varies, we must have $\frac{dh}{d\theta} = 0$.
Substituting $\frac{dh}{d\theta} = 0$ into the differentiated equation yields:
$-20h\cos 2\theta + 30\sin 2\theta = 0$
$20h\cos 2\theta = 30\sin 2\theta$
$h = \frac{30\sin 2\theta}{20\cos 2\theta} = \frac{3}{2}\tan 2\theta$

3. Substitute the expression for $h$ back into the original equation.

Substitute $h = \frac{3}{2}\tan 2\theta$ into the original equation from (a):
$\left(\frac{3}{2}\tan 2\theta\right)^2 - 10\left(\frac{3}{2}\tan 2\theta\right)\sin 2\theta - 15\cos 2\theta - 15 = 0$
$\frac{9}{4}\tan^2 2\theta - 15\tan 2\theta\sin 2\theta - 15\cos 2\theta - 15 = 0$

4. Multiply the equation by a constant to obtain the target form.

Multiply the entire equation by $\frac{4}{3}$ to clear the fraction and match the required coefficients:
$\frac{4}{3}\left(\frac{9}{4}\tan^2 2\theta\right) - \frac{4}{3}(15\tan 2\theta\sin 2\theta) - \frac{4}{3}(15\cos 2\theta) - \frac{4}{3}(15) = 0$
$3\tan^2 2\theta - 20\sin 2\theta\tan 2\theta - 20\cos 2\theta - 20 = 0 \text{ (shown)}$

1. Solve the equation from part (b) to find exact trigonometric ratios for $2\theta$.

$3\tan^2 2\theta - 20\sin 2\theta\tan 2\theta - 20\cos 2\theta - 20 = 0$
Rewrite $\tan 2\theta$ in terms of sine and cosine:
$3(\sec^2 2\theta - 1) - 20\left(\frac{\sin^2 2\theta}{\cos 2\theta} + \cos 2\theta\right) - 20 = 0$
$3\sec^2 2\theta - 3 - 20\left(\frac{\sin^2 2\theta + \cos^2 2\theta}{\cos 2\theta}\right) - 20 = 0$
$3\sec^2 2\theta - 20\sec 2\theta - 23 = 0$
Factorise the quadratic in $\sec 2\theta$:
$(3\sec 2\theta - 23)(\sec 2\theta + 1) = 0$
Since the angle of projection $\theta$ must be acute, $\sec 2\theta \neq -1$ (which would imply projecting vertically, giving $h=0$). Thus, $\sec 2\theta = \frac{23}{3}$.
This gives $\cos 2\theta = \frac{3}{23}$ and $\sin 2\theta = \sqrt{1 - \left(\frac{3}{23}\right)^2} = \frac{2\sqrt{130}}{23}$.

2. Substitute the trigonometric values into the equation from part (a) to find $h$.

$h^2 - 10h\sin 2\theta - 15\cos 2\theta - 15 = 0$
Substitute $\sin 2\theta = \frac{2\sqrt{130}}{23}$ and $\cos 2\theta = \frac{3}{23}$:
$h^2 - 10h\left(\frac{2\sqrt{130}}{23}\right) - 15\left(\frac{3}{23}\right) - 15 = 0$
Multiply through by $23$:
$23h^2 - 20\sqrt{130}h - 390 = 0$
Use the quadratic formula, keeping in mind that $h > 0$:
$h = \frac{20\sqrt{130} + \sqrt{(20\sqrt{130})^2 - 4(23)(-390)}}{2(23)}$
$h = \frac{20\sqrt{130} + \sqrt{52000 + 35880}}{46}$
$h = \frac{20\sqrt{130} + \sqrt{87880}}{46} = \frac{20\sqrt{130} + 26\sqrt{130}}{46}$
$h = \frac{46\sqrt{130}}{46} = \sqrt{130}$