Question 13 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 13 - 2016 - Paper 1

To find when the tide is increasing at the fastest rate, we need to analyze the first derivative of the function. The rate of change of the tide, $\frac{dx}{dt}$, is given by: $\frac{dx}{dt} = -4 \cdot \frac{4\pi}{25} \sin \left( \frac{4\pi}{25} t \right)$ The tide increases when $\sin \left( \frac{4\pi}{25} t \right)$ is positive, which occurs between $0$ and $\pi$. The fastest rate (maximum value) occurs at $\frac{4\pi}{25} t = \frac{\pi}{2}$. Therefore: $t = \frac{25}{8} \text{ hours after 2 am} = 3:30 ext{ am}$

The vertical component of the initial velocity is $u \sin \theta$. The maximum height is reached when the vertical velocity is zero, which can be found using: $v^2 = u^2 + 2as$ Here, $v = 0$, $a = -10$ (gravity), hence: $0 = (u \sin \theta)^2 - 20h \\ h = \frac{(u \sin \theta)^2}{20}$

First, find the time to reach the highest point, given by: $t = \frac{u \sin(30°)}{10} = \frac{30 \cdot 0.5}{10} = 1.5 ext{ seconds}$ At this instant, we find the height using: $y = 20 + 30 \cdot 0.5 \cdot 1.5 - 5 \cdot (1.5)^2$ Evaluate this to get: $y = 20 + 22.5 - 11.25 = \frac{125}{4} ext{ m}$

The ball rebounds vertically after hitting the wall with speed 10 m/s. Using: $h = \frac{1}{2} g t^2$ where height $h$ is $\frac{125}{4}$ m - 20 m = $\frac{125 - 80}{4} = \frac{45}{4}$. Solving: $\frac{45}{4} = \frac{1}{2} \cdot 10 imes t^2 \\ t^2 = \frac{90}{10} \\ t = 3\text{ seconds}$

Since the horizontal speed is constant, the horizontal distance traveled in the rebound phase is: Horizontal speed = 10 m/s for 3 seconds: $ext{Distance} = 10 \cdot 3 = 30 ext{ m}$

To show that CMDE is a cyclic quadrilateral, we note:

• Angles subtended by the same chord (CD) are equal. Hence, $\angle CED = \angle CMD$. Therefore, CMDE lies on a circle.

Since CMDE is cyclic, we have: $\angle CED + \angle CMD = 180° \\\\ ext{and }\angle MDF = 90° ext{ as angles in a semicircle.}$ Thus, MF is perpendicular to AB.