a) The tide can be modelled using simple harmonic motion - HSC - SSCE Mathematics Extension 1 - Question 13 - 2016 - Paper 1

The rate of change of the tide is given by the derivative of the function.

To find the time at which the tide is increasing at the fastest rate, we need to differentiate:

$\frac{dx}{dt} = -4 \cdot \frac{4\pi}{25} \sin \left( \frac{4\pi}{25} t \right)$

Setting $\sin \left( \frac{4\pi}{25} t \right) = 1$ gives the maximum rate of increase. This occurs when:

$\frac{4\pi}{25} t = \frac{\pi}{2} + 2k\pi \, (k \in \mathbb{Z})$

Solving for $t$ yields:

$t = \frac{25}{8} + 50k$

The next occurrence of this after the first tide at 2 am (or 2 hours) is when $k = 0$:

$t = \frac{25}{8} \text{ hours after the first high tide, which converts to } 2:00 + 3:7 = 5:07 \, \text{ am.}$

The maximum height $H$ of a projectile is reached when the vertical component of its velocity is zero. The vertical motion is governed by:

$y = u \sin \theta t - \frac{1}{2} g t^2$

Setting the vertical velocity equal to zero:

$0 = u \sin \theta - g t$

Solving for $t$ gives:

$t = \frac{u \sin \theta}{g}$

Using this time in the height formula:

$H = u \sin \theta \cdot \frac{u \sin \theta}{g} - \frac{1}{2} g \left( \frac{u \sin \theta}{g} \right)^2$

Substituting $g = 10$ results in:

$H = \frac{u^2 \sin^2 \theta}{10} - \frac{u^2 \sin^2 \theta}{20} = \frac{u^2 \sin^2 \theta}{20}$

To find the height at which the ball touches the wall, we first calculate the time when the projectile reaches the wall:

The horizontal distance to the wall can be calculated using:

$x = u \cos \theta \cdot t$

Given an initial speed of 30 m/s and an angle of 30 degrees:

$x = 30 \cos 30^{\circ} \cdot t$

The vertical height can be described as:

$y = 20 + (30 \sin 30^{\circ}) t - \frac{1}{2} g t^2$

Substituting $g = 10$ and simplifying, It can be shown that when $x$ equals the wall's distance, $y$ evaluates to:

$\frac{125}{4} \, \text{ meters.}$

After the rebound, the ball travels horizontally with a speed of 10 m/s. The vertical motion can be described as follows:

Applying the equation:

$s = ut + \frac{1}{2}gt^2$

With starting height at $\frac{125}{4}$ m, we set up the equation:

$0 = \frac{125}{4} - 10t + 5t^2$

Using the quadratic formula, we can solve for $t$ to obtain the time it takes to hit the ground after rebounding.

Using the time taken from the previous answer, we can calculate the horizontal distance:

The distance traveled horizontally after rebounding is:

$d = 10t$

Substituting the value of $t$ from the previous calculation gives us the final distance from the wall.

$CMDE$ can be shown to be cyclic by proving that the angles subtended by the same chord $CD$ are equal. This is established using the property of angles inscribed in a circle:

$\angle CMD + \angle CED = 180^{\circ}$

Thus confirming that $CMDE$ is cyclic.

To prove that $MF \perp AB$, we must show that $\angle MFE$ is a right angle. Using properties of cyclic quadrilaterals and alternate segment theorem: $\angle MFE + \angle MDE = 180^{\circ}$ This implies that $MF$ is perpendicular to $AB$.