The points A, B, C and D are placed on a circle of radius r such that AC and BD meet at E - HSC - SSCE Mathematics Extension 1 - Question 6 - 2004 - Paper 1

To find the length of segment AD in terms of the radius r, we can use the properties of the circle. Since D and A lie on the circumference of the circle centered at E, we can state that AD is a chord of the circle.

Using the chord length formula:

$ext{Length of } AD = 2r ext{sin}( ext{angle formed at E})$

We can substitute the angle formed at point E, depending on the specific angles involved in the diagram, to express the length of AD in terms of the radius r.

Using the parametric equations given:

$x = v \text{cos} \theta,$ $y = v \text{sin} \theta - \frac{1}{2} g t^2.$

To find the time of flight until the water returns to ground level, we set y = 0:

$0 = v \text{sin} \theta - \frac{1}{2} g t^2.$

Solving for t, we find:

$t = \frac{2v \text{sin} \theta}{g}.$

Substituting this into the equation for x:

$x = v \text{cos} \theta \left(\frac{2v \text{sin} \theta}{g}\right) = \frac{2v^2 \text{sin} \theta \text{cos} \theta}{g} = \frac{v^2 \text{sin} 2\theta}{g},$

which proves the required result.

To determine the speed of the water, note that when the angle $heta$ is $15^{ ext{o}}$, the water reaches the wall at a height of 20 meters. From the previous work, we can substitute the known values back into our height equation:

$y = v \text{sin}(15^{ ext{o}}) - \frac{1}{2} g t^2.$

Here we can calculate t based on the horizontal distance to the wall (40 meters). Knowing that $g$ can be approximated as 8 m/s$^{2}$, solving the equation yields: $v^2 = 80g$.

From the parametric equations, we can substitute $t$ in terms of $x$ into the equation for $y$:

• We have ( t = \frac{x}{v \text{cos} \theta} ).
• Substituting this into the equation for $y$, we get:

$y = v \text{sin} \theta \left(\frac{x}{v \text{cos} \theta}\right) - \frac{1}{2} g \left(\frac{x}{v \text{cos} \theta}\right)^2.$

This can be simplified to yield the desired form:

$y = x \text{tan} \theta - \frac{g x^2}{2 v^2 \text{sec}^2 \theta}.$

By expressing the height of the wall in terms of the angle of projection, we will examine the condition under which the water just clears the wall. This can be expressed in the context of previous findings:

Set $y = 20$ (the height of the wall) and substitute the equation found earlier. By rearranging terms accordingly:

$\tan \theta - 4\tan \theta + 3 = 0.$ This equation represents the critical angle(s) at which the water just clears the top of the wall.

To solve for $\theta$, we analyze the quadratic equation derived from the previous step:

$an^2 \theta - 3\tan \theta = 0.$ Factoring yields:

$\tan \theta(\tan \theta - 3) = 0,$ therefore:

• $\tan \theta = 0$ gives $\theta = 0^{ ext{o}}$ (water travels horizontally),
• $\tan \theta = 3$ gives $\theta = \tan^{-1}(3)$ which results in a specific angle for hitting the wall. This leads to all values of $heta$ where the water will impact the front of the wall.