The diagram shows a shot put being projected with a velocity $v\text{ ms}^{-1}$ from the point $O$ at an angle $\theta$ made with the horizontal. The point $O$ is $1.5\text{m}$ 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
$y = x\tan\theta - \frac{gx^2}{2v^2\cos^2\theta},$
where $g\text{ ms}^{-2}$ is the acceleration due to gravity.
The constant $g$ is taken to be $10$ and the distance between $A$ and $B$ is denoted by $h\text{ m}$.

1. State the coordinates of the landing point $B$ to substitute into the trajectory equation.

Since $O$ is $1.5\text{ m}$ above $A$ and the horizontal distance between $A$ and $B$ is $h\text{ m}$, the coordinates of point $B$ on the $x$-$y$ plane are $(h, -1.5)$.

2. Substitute the coordinates of $B$ and the given values into the trajectory equation.

Substitute $x = h$, $y = -1.5$, $v = 10$, and $g = 10$ into the trajectory equation:
$-1.5 = h\tan\theta - \frac{10h^2}{2(10)^2\cos^2\theta}$
$-1.5 = h\tan\theta - \frac{h^2}{20\cos^2\theta}$

3. Rearrange the equation and express $\tan\theta$ in terms of sine and cosine.

Multiply both sides by $20\cos^2\theta$:
$-30\cos^2\theta = 20h\left(\frac{\sin\theta}{\cos\theta}\right)\cos^2\theta - h^2$
$-30\cos^2\theta = 20h\sin\theta\cos\theta - h^2$
$h^2 - 20h\sin\theta\cos\theta - 30\cos^2\theta = 0$

4. Apply double angle identities to obtain the target equation.

Use the double angle identities $2\sin\theta\cos\theta = \sin 2\theta$ and $2\cos^2\theta = \cos 2\theta + 1$:
$h^2 - 10h(2\sin\theta\cos\theta) - 15(2\cos^2\theta) = 0$
$h^2 - 10h\sin 2\theta - 15(\cos 2\theta + 1) = 0$
$h^2 - 10h\sin 2\theta - 15\cos 2\theta - 15 = 0 \quad \text{(Shown)}$

1. Differentiate the given equation implicitly with respect to $\theta$.

$\frac{\mathrm{d}}{\mathrm{d}\theta} (h^2 - 10h\sin 2\theta - 15\cos 2\theta - 15) = 0$
$2h \frac{\mathrm{d}h}{\mathrm{d}\theta} - \left( 10\frac{\mathrm{d}h}{\mathrm{d}\theta}\sin 2\theta + 10h(2\cos 2\theta) \right) - 15(-2\sin 2\theta) = 0$
$2h \frac{\mathrm{d}h}{\mathrm{d}\theta} - 10\frac{\mathrm{d}h}{\mathrm{d}\theta}\sin 2\theta - 20h\cos 2\theta + 30\sin 2\theta = 0$

2. Apply the condition for a stationary value of $h$.

For a stationary value of $h$, we set $\frac{\mathrm{d}h}{\mathrm{d}\theta} = 0$:
$0 - 0 - 20h\cos 2\theta + 30\sin 2\theta = 0$
$-20h\cos 2\theta + 30\sin 2\theta = 0$

3. Solve for $h$ in terms of $\theta$.

$20h\cos 2\theta = 30\sin 2\theta$
$h = \frac{30\sin 2\theta}{20\cos 2\theta}$
$h = \frac{3}{2}\tan 2\theta$

4. Substitute the expression for $h$ back into the original equation to eliminate $h$.

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

5. Simplify the equation to match the required form.

Multiply the entire equation by $\frac{4}{3}$:
$\frac{4}{3} \times \left(\frac{9}{4}\tan^2 2\theta - 15\tan 2\theta\sin 2\theta - 15\cos 2\theta - 15\right) = 0$
$3\tan^2 2\theta - 20\sin 2\theta\tan 2\theta - 20\cos 2\theta - 20 = 0 \quad \text{(Shown)}$

1. Convert the stationary condition into a polynomial equation in terms of $\cos 2\theta$.

$3\tan^2 2\theta - 20\sin 2\theta\tan 2\theta - 20\cos 2\theta - 20 = 0$
Substitute $\tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta}$ and multiply by $\cos^2 2\theta$:
$3\sin^2 2\theta - 20\sin^2 2\theta\cos 2\theta - 20\cos^3 2\theta - 20\cos^2 2\theta = 0$
Substitute $\sin^2 2\theta = 1 - \cos^2 2\theta$:
$3(1 - \cos^2 2\theta) - 20\cos 2\theta(1 - \cos^2 2\theta) - 20\cos^3 2\theta - 20\cos^2 2\theta = 0$
$3 - 3\cos^2 2\theta - 20\cos 2\theta + 20\cos^3 2\theta - 20\cos^3 2\theta - 20\cos^2 2\theta = 0$
$23\cos^2 2\theta + 20\cos 2\theta - 3 = 0$

2. Solve for $\cos 2\theta$ and find the corresponding $\sin 2\theta$ for the valid projectile angle.

$(23\cos 2\theta - 3)(\cos 2\theta + 1) = 0$
$\cos 2\theta = \frac{3}{23} \quad \text{or} \quad \cos 2\theta = -1$
Since $\cos 2\theta = -1 \implies \theta = 90^\circ$ (vertical throw resulting in a minimal $h=0$), we take $\cos 2\theta = \frac{3}{23}$.
For an acute angle of projection $\theta$, $\sin 2\theta > 0$:
$\sin 2\theta = \sqrt{1 - \left(\frac{3}{23}\right)^2} = \frac{2\sqrt{130}}{23}$

3. Substitute $\cos 2\theta$ and $\sin 2\theta$ into the trajectory equation from part (a) to form a quadratic equation in $h$.

$h^2 - 10h\left(\frac{2\sqrt{130}}{23}\right) - 15\left(\frac{3}{23}\right) - 15 = 0$
$h^2 - \frac{20\sqrt{130}}{23} h - \frac{390}{23} = 0$

4. Solve the quadratic equation for $h$ to find the stationary value.

Using the quadratic formula:
$h = \frac{ \frac{20\sqrt{130}}{23} \pm \sqrt{ \left(-\frac{20\sqrt{130}}{23}\right)^2 - 4(1)\left(-\frac{390}{23}\right) } }{ 2 }$
$h = \frac{ \frac{20\sqrt{130}}{23} \pm \sqrt{ \frac{52000}{529} + \frac{35880}{529} } }{ 2 }$
$h = \frac{ \frac{20\sqrt{130}}{23} \pm \frac{\sqrt{87880}}{23} }{ 2 }$
Note that $\sqrt{87880} = \sqrt{676 \times 130} = 26\sqrt{130}$:
$h = \frac{10\sqrt{130} \pm 13\sqrt{130}}{23}$
Since the landing distance $h$ must be positive, reject the negative root:
$h = \frac{23\sqrt{130}}{23} = \sqrt{130} \approx 11.4 \text{ m}$