Answer:
a) [tex]\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j[/tex], b) [tex]\theta = \frac{\pi}{4}[/tex], c) [tex]y_{max} = 84.375\,m[/tex], [tex]t = 3.75\,s[/tex].
Step-by-step explanation:
a) The function in terms of time and the inital angle measured from the horizontal is:
[tex]\vec r (t) = [(v_{o}\cdot \cos \theta)\cdot t]\cdot i + \left[(v_{o}\cdot \sin \theta)\cdot t -\frac{1}{2}\cdot g \cdot t^{2} \right]\cdot j[/tex]
The particular expression for the cannonball is:
[tex]\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j[/tex]
b) The components of the position of the cannonball before hitting the ground is:
[tex]x = (90\cdot \cos \theta)\cdot t[/tex]
[tex]0 = 90\cdot \sin \theta - 6\cdot t[/tex]
After a quick substitution and some algebraic and trigonometric handling, the following expression is found:
[tex]0 = 90\cdot \sin \theta - 6\cdot \left(\frac{x}{90\cdot \cos \theta} \right)[/tex]
[tex]0 = 8100\cdot \sin \theta \cdot \cos \theta - 6\cdot x[/tex]
[tex]0 = 4050\cdot \sin 2\theta - 6\cdot x[/tex]
[tex]6\cdot x = 4050\cdot \sin 2\theta[/tex]
[tex]x = 675\cdot \sin 2\theta[/tex]
The angle for a maximum horizontal distance is determined by deriving the function, equalizing the resulting formula to zero and finding the angle:
[tex]\frac{dx}{d\theta} = 1350\cdot \cos 2\theta[/tex]
[tex]1350\cdot \cos 2\theta = 0[/tex]
[tex]\cos 2\theta = 0[/tex]
[tex]2\theta = \frac{\pi}{2}[/tex]
[tex]\theta = \frac{\pi}{4}[/tex]
Now, it is required to demonstrate that critical point leads to a maximum. The second derivative is:
[tex]\frac{d^{2}x}{d\theta^{2}} = -2700\cdot \sin 2\theta[/tex]
[tex]\frac{d^{2}x}{d\theta^{2}} = -2700[/tex]
Which demonstrates the existence of the maximum associated with the critical point found before.
c) The equation for the vertical component of position is:
[tex]y = 45\cdot t - 6\cdot t^{2}[/tex]
The maximum height can be found by deriving the previous expression, which is equalized to zero and critical values are found afterwards:
[tex]\frac{dy}{dt} = 45 - 12\cdot t[/tex]
[tex]45-12\cdot t = 0[/tex]
[tex]t = \frac{45}{12}[/tex]
[tex]t = 3.75\,s[/tex]
Now, the second derivative is used to check if such solution leads to a maximum:
[tex]\frac{d^{2}y}{dt^{2}} = -12[/tex]
Which demonstrates the assumption.
The maximum height reached by the cannonball is:
[tex]y_{max} = 45\cdot (3.75\,s)-6\cdot (3.75\,s)^{2}[/tex]
[tex]y_{max} = 84.375\,m[/tex]
