Let's fire a cannon. Recall the general equation for projectile motion. The height [math]y[/math] metres of the ball whose horizontal distance is [math]x[/math] metres away is given by the equation[center][br][math]y=ax^2+bx+c,[/math][br][/center]where [math]a=-\frac{g}{2v_0^2\left(\cos\theta\right)^2}[/math], [math]b=\tan\theta[/math], and [math]c=y_0[/math].[br][br]Here, [math]v_0\ \mathrm{m}/\mathrm{s}[/math] denotes the initial speed, [math]y_0\ \text{m}[/math] denotes the initial height, [math]g \approx 9.81\ \text{m}/\text{s}^2[/math] denotes the [b]gravitational constant[/b] on earth, and [math]\theta[/math] denotes the angle of elevation.
The cannon fires from ground level at a initial speed of [math]250\ \text{m}/\text{s}[/math].
[b]Question 1.[/b] Write down the values of [math]v_0[/math] and [math]y_0[/math].
[math]v_0=250[/math], [math]y_0=0[/math]
Create a GeoGebra simulation given your values of [math]v[/math] and [math]y_0[/math] in the previous part by entering the following commands into the input bar. Press ENTER after each input.[br][list=1][*]T = Slider(0.1, 89.9, 0.1)[/*][*]S = T * pi / 180[br][/*][*]g = 9.81[/*][*]a = g / (200 * (cos(S))^2)[/*][*]b = tan(S)[/*][*]c = 0[/*][*]y = a*x^2 + b*x + c[/*][/list]
[b]Question 2. [/b]Fix [math]\theta=30^{\circ}[/math]. What is the horizontal distance that the cannonball travels?
Approximately [math]5520\ \text{m}[/math]
[b]Question 3. [/b]Find a different value of [math]\theta[/math] that gives the same horizontal distance.
[math]\theta=60^{\circ}[/math]
[b]Question 4. [/b]Find the value of [math]\theta[/math] that maximises the horizontal distance.
[math]\theta=45^{\circ}[/math]
[b]Bonus Question.[/b] Why does this happen?
[math]\sin(2\theta)[/math] is maximised at [math]1[/math] when [math]2\theta = {90}^\circ\ \Leftrightarrow\ \theta = {45}^\circ[/math].