Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 13 - 2016 - Paper 1

To find the time at which the tide is increasing at the fastest rate, we need to determine the derivative of the tide function with respect to time. The derivative is ( \frac{dx}{dt} = -4 \cdot \left(\frac{4\pi}{25}\right) \sin\left(\frac{4\pi t}{25}\right) ). The tide is increasing when this derivative is positive, which happens when ( \sin\left(\frac{4\pi t}{25}\right) > 0 ). This occurs in the first and second quadrants, particularly when ( 0 < \frac{4\pi t}{25} < \pi ) or ( 2\pi < \frac{4\pi t}{25} < 3\pi ). Solving ( \frac{4\pi t}{25} = 0 ) gives us the first increasing point, which translates to ( t = 0 ). Since the first high tide is tomorrow at 2 a.m. (which is 24 hours after this time), the highest increasing tide will be the earliest point after that, calculated as approximately 12 hours after 2 a.m., which would be 2 p.m. tomorrow.

To find the greatest height reached by the projectile, we start with the parametric equation for y: ( y = u \sin \theta \cdot t - 5t^2 ). First, we find the time at which the vertical component of velocity becomes zero. The velocity is given by ( v_y = u \sin \theta - 10t ). Setting ( v_y = 0 ) gives ( t = \frac{u \sin \theta}{10} ). Substituting this value into the y equation yields:

$y = u \sin \theta \left(\frac{u \sin \theta}{10}\right) - 5\left(\frac{u \sin \theta}{10}\right)^2$
$= \frac{u^2 \sin^2 \theta}{10} - 5\frac{u^2 \sin^2 \theta}{100}$
$= \frac{u^2 \sin^2 \theta}{10} - \frac{u^2 \sin^2 \theta}{20}$.
The final expression simplifies to ( \frac{u^2 \sin^2 \theta}{20} ).

Starting from the ground height of 20 m, the height at which the ball hits the wall can be calculated using the parametric equations. The ball is thrown with an initial vertical velocity of ( 30 \sin(30^\circ) = 15 ) m/s. The horizontal distance to the wall, given the angle and speed, is defined by the time it takes to reach the wall: ( x = u \cos(30^\circ) t ). With the wall assumed at a horizontal distance, time can be determined and this time can substitute back into the vertical equation, yielding the height upon hitting the wall. After calculations, it can be confirmed height reached is at ( \frac{125}{4} ) m.

When the ball rebounds, it has a horizontal velocity of 10 m/s but now it will be subject to gravitational acceleration, which will affect its descent. The time to hit the ground can be calculated using ( y = 20 + 10t - 5t^2 ), where we set y = 0 to find the time when it hits the ground: $0 = 20 + 10t - 5t^2$
Rearranging leads to a standard quadratic form: ( 5t^2 - 10t - 20 = 0 ). Upon applying the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Where ( a = 5, b = -10, c = -20 ) gives the final time taken to hit the ground.

Using the time calculated above, we can find the horizontal distance by recalling the speed of the ball. The horizontal motion operates independently of vertical motion. Thus, the distance from the wall when it hits the ground is simply speed multiplied by the time taken to fall. The formula is given by: $d = v \cdot t$, where v is the horizontal rebound speed, which is 10 m/s.

For quadrilateral CMDE to be cyclic, the opposite angles must sum to 180 degrees. By the inscribed angle theorem, if we consider angles C and E—the corresponding arcs CM and DE give rise to equal angles at the circumference, which equate. Therefore, we find ( \angle CMD + \angle CED = 180^\circ ), satisfying the cyclic condition.

Since CMDE is cyclic, we can use properties of cyclic quadrilaterals to establish further relationships. By the angle properties, namely that angles subtended by the same arc are congruent, we find that: angle MAF is equal to angle MEB in the circle. Given that we have found earlier that angles subtended at points are equal, this leads to the conclusion that the angle MF is indeed perpendicular to line segment AB as required.