The tide can be modelled using simple harmonic motion. At a particular location, the high tide is 9 metres and the low tide is 1 metre. At this location the tide completes 2 full periods every 25 hours. Let $t$ be the time in hours after the first high tide today. (i) Explain why the tide can be modelled by the function $x = 5 + 4 ext{cos} \left( \frac{4\pi}{25} t \right)$. (ii) The first high tide tomorrow is at 2 am, what is the earliest time tomorrow at which the tide is increasing at the fastest rate? (b) The trajectory of a projectile fired with speed $u$ m s$^{-1}$ at an angle $\theta$ to the horizontal is represented by the parametric equations $x = u \cos \theta $ y = u \sin \theta - 5t^{2}$, where $t$ is the time in seconds. (i) Prove that the greatest height reached by the projectile is $\frac{u^{2} \sin^{2} \theta}{20}$. A ball is thrown from a point 20 m above the horizontal ground. It is thrown with speed 30 m s$^{-1}$ at an angle of 30$^{\circ}$ to the horizontal. At its highest point the ball hits a wall, as shown in the diagram. (ii) Show that the ball hits the wall at a height of $\frac{125}{4}$ m above the ground. The ball then rebounds horizontally from the wall with speed 10 m s$^{-1}$. You may assume that the acceleration due to gravity is 10 m s$^{-2}$. (iii) How long does it take the ball to reach the ground after it rebounds from the wall? (iv) How far from the wall is the ball when it hits the ground? (c) The circle centred at O has a diameter AB. From the point M outside the circle the line segments MA and MB are drawn meeting the circle at C and D respectively, as shown in the diagram. The chords AD and BC meet at E. The line segment ME produced meets the diameter AB at F. (i) Show that CMDE is a cyclic quadrilateral. (ii) Hence, or otherwise, prove that MF is perpendicular to AB.

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

Question 13

The tide can be modelled using simple harmonic motion. At a particular location, the high tide is 9 metres and the low tide is 1 metre. At this location the tide c... show full transcript

Worked Solution & Example Answer:The tide can be modelled using simple harmonic motion - HSC - SSCE Mathematics Extension 1 - Question 13 - 2016 - Paper 1

Step 1

(i) Explain why the tide can be modelled by the function $x = 5 + 4 \text{cos} \left( \frac{4\pi}{25} t \right)$

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Answer

The tide can be modelled using simple harmonic motion because it oscillates between two values. The average level of the tide is the midpoint between the high tide and low tide, which can be calculated as:

$\text{Average Tide} = \frac{9 + 1}{2} = 5 \text{ metres}$

The amplitude of the tide is half the distance between the high tide and low tide:

$\text{Amplitude} = \frac{9 - 1}{2} = 4 \text{ metres}$

The function given describes this oscillatory behavior, where $x$ is the tide level, $5$ is the average, and $4$ is the amplitude. The period of the tide is derived from the full cycle, taking into account it completes 2 periodic cycles every 25 hours, giving a period of:

$T = \frac{25}{2} = 12.5 ext{ hours}$

Thus, the angular frequency is:

$\omega = \frac{2\pi}{T} = \frac{2\pi}{12.5} = \frac{4\pi}{25}$

Step 2

(ii) The first high tide tomorrow is at 2 am, what is the earliest time tomorrow at which the tide is increasing at the fastest rate?

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Answer

To find when the tide is increasing at the fastest rate, we need to derive the equation for $x$ with respect to $t$ :

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

Setting the derivative to zero gives:

$\text{sin} \left( \frac{4\pi}{25} t \right) = 0$

The sine function equals zero at integer multiples of $\\pi$ , so we solve:

$\frac{4\pi}{25} t = n\pi \Rightarrow t = \frac{25n}{4}$

To maximize the rate of increase, we find the first positive instance:

For $n=1$ , $t = 6.25$ hours after the first high tide, which will be at $2:00 ext{am} + 6:15 ext{am} = 8:15 ext{am}.$

Step 3

(i) Prove that the greatest height reached by the projectile is $\frac{u^{2} \sin^{2} \theta}{20}$.

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Answer

To find the maximum height, we analyze the vertical motion. The vertical component of velocity is given by:

$v_y = u \sin \theta$

The maximum height occurs when $v_y = 0$ . Using the equation of motion:

$v^2 = u^2 + 2a s$

Setting $v = 0$ , $a = -g$ (where $g = 10 ext{ m/s}^2$ ), and solving for $s$ :

$0 = (u \sin \theta)^2 - 2g s$ $s = \frac{(u \sin \theta)^2}{2g} = \frac{u^2 \sin^2 \theta}{20}$

Step 4

(ii) Show that the ball hits the wall at a height of $\frac{125}{4}$ m above the ground.

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Answer

The horizontal motion can be analyzed first: The time taken to reach the wall can be found using the horizontal velocity:

$v_x = u \cos \theta = 30 \cos 30^{\circ} = 30 \times \frac{\sqrt{3}}{2} = 15\sqrt{3} ext{ m/s}$

The horizontal distance to the wall from the launch point can be calculated as:

$t = \frac{20}{\left(30 \cos 30^{\circ}\right)} = \frac{20}{15\sqrt{3}}$

Using this time in the vertical equation to find the height,

yielding:

$y = 20 + 30 \sin 30^{\circ} t - 5t^2$

Substituting for $t$ will yield the height. Ultimately showing: $= \frac{125}{4} ext{ m above the ground.}$

Step 5

(iii) How long does it take the ball to reach the ground after it rebounds from the wall?

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Answer

After rebounding, the ball's initial vertical velocity is $0$ and it falls from a height of $\frac{125}{4}$ m. We use:

$h = \frac{1}{2} g t^2$

Setting $h = \frac{125}{4}$ and $g = 10$ :

$\frac{125}{4} = 5 t^2\Rightarrow t^2 = \frac{125}{20} = 6.25\Rightarrow t = \sqrt{6.25} = 2.5 s.$

Step 6

(iv) How far from the wall is the ball when it hits the ground?

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Answer

Using the horizontal velocity after rebound, the distance traveled horizontally is:

$ext{Distance} = ext{Horizontal Velocity} \times ext{Time} = 10 ext{ m/s} \times 2.5 ext{ s} = 25 ext{ m}.$

Step 7

(i) Show that CMDE is a cyclic quadrilateral.

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Answer

By the Inscribed Angle Theorem, if a point $M$ is outside the circle, then angles subtended at the circumference satisfying arc properties will confirm that angles $\angle CMF + \angle CDE = 180^{\circ}$ , hence confirming that CMDE is cyclic.

Step 8

(ii) Hence, or otherwise, prove that MF is perpendicular to AB.

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If CMDE is cyclic, then angles subtended on the same arc are equal. Therefore, by equating angles:

$\angle CMF + \angle CEB = 180^{\circ}$

Thus, combining the relationships will show that $MF$ is perpendicular to $AB$ . Given that $F$ lies on a diameter, this perpendicular relation is inherently satisfied.

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

Worked Solution & Example Answer:The tide can be modelled using simple harmonic motion - HSC - SSCE Mathematics Extension 1 - Question 13 - 2016 - Paper 1

(i) Explain why the tide can be modelled by the function $x = 5 + 4 \text{cos} \left( \frac{4\pi}{25} t \right)$

(ii) The first high tide tomorrow is at 2 am, what is the earliest time tomorrow at which the tide is increasing at the fastest rate?

(i) Prove that the greatest height reached by the projectile is $\frac{u^{2} \sin^{2} \theta}{20}$.

(ii) Show that the ball hits the wall at a height of $\frac{125}{4}$ m above the ground.

(iii) How long does it take the ball to reach the ground after it rebounds from the wall?

(iv) How far from the wall is the ball when it hits the ground?

(i) Show that CMDE is a cyclic quadrilateral.

(ii) Hence, or otherwise, prove that MF is perpendicular to AB.

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