Answer:
a)
, b)
, c)
,
.
Step-by-step explanation:
a) The function in terms of time and the inital angle measured from the horizontal is:
![\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](https://tex.z-dn.net/?f=%5Cvec%20r%20%28t%29%20%3D%20%5B%28v_%7Bo%7D%5Ccdot%20%5Ccos%20%5Ctheta%29%5Ccdot%20t%5D%5Ccdot%20i%20%2B%20%5Cleft%5B%28v_%7Bo%7D%5Ccdot%20%5Csin%20%5Ctheta%29%5Ccdot%20t%20-%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20g%20%5Ccdot%20t%5E%7B2%7D%20%5Cright%5D%5Ccdot%20j)
The particular expression for the cannonball is:
![\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](https://tex.z-dn.net/?f=%5Cvec%20r%20%28t%29%20%3D%20%5Cleft%5B%2890%5Ccdot%20%5Ccos%20%5Ctheta%29%5Ccdot%20t%20%5Cright%5D%5Ccdot%20i%20%2B%20%5Cleft%5B%2890%5Ccdot%20%5Csin%20%5Ctheta%29%5Ccdot%20t%20-6%5Ccdot%20t%5E%7B2%7D%20%5Cright%5D%5Ccdot%20j)
b) The components of the position of the cannonball before hitting the ground is:
![x = (90\cdot \cos \theta)\cdot t](https://tex.z-dn.net/?f=x%20%3D%20%2890%5Ccdot%20%5Ccos%20%5Ctheta%29%5Ccdot%20t)
![0 = 90\cdot \sin \theta - 6\cdot t](https://tex.z-dn.net/?f=0%20%3D%2090%5Ccdot%20%5Csin%20%5Ctheta%20-%206%5Ccdot%20t)
After a quick substitution and some algebraic and trigonometric handling, the following expression is found:
![0 = 90\cdot \sin \theta - 6\cdot \left(\frac{x}{90\cdot \cos \theta} \right)](https://tex.z-dn.net/?f=0%20%3D%2090%5Ccdot%20%5Csin%20%5Ctheta%20-%206%5Ccdot%20%5Cleft%28%5Cfrac%7Bx%7D%7B90%5Ccdot%20%5Ccos%20%5Ctheta%7D%20%20%5Cright%29)
![0 = 8100\cdot \sin \theta \cdot \cos \theta - 6\cdot x](https://tex.z-dn.net/?f=0%20%3D%208100%5Ccdot%20%5Csin%20%5Ctheta%20%5Ccdot%20%5Ccos%20%5Ctheta%20-%206%5Ccdot%20x)
![0 = 4050\cdot \sin 2\theta - 6\cdot x](https://tex.z-dn.net/?f=0%20%3D%204050%5Ccdot%20%5Csin%202%5Ctheta%20-%206%5Ccdot%20x)
![6\cdot x = 4050\cdot \sin 2\theta](https://tex.z-dn.net/?f=6%5Ccdot%20x%20%3D%204050%5Ccdot%20%5Csin%202%5Ctheta)
![x = 675\cdot \sin 2\theta](https://tex.z-dn.net/?f=x%20%3D%20675%5Ccdot%20%5Csin%202%5Ctheta)
The angle for a maximum horizontal distance is determined by deriving the function, equalizing the resulting formula to zero and finding the angle:
![\frac{dx}{d\theta} = 1350\cdot \cos 2\theta](https://tex.z-dn.net/?f=%5Cfrac%7Bdx%7D%7Bd%5Ctheta%7D%20%3D%201350%5Ccdot%20%5Ccos%202%5Ctheta)
![1350\cdot \cos 2\theta = 0](https://tex.z-dn.net/?f=1350%5Ccdot%20%5Ccos%202%5Ctheta%20%3D%200)
![\cos 2\theta = 0](https://tex.z-dn.net/?f=%5Ccos%202%5Ctheta%20%3D%200)
![2\theta = \frac{\pi}{2}](https://tex.z-dn.net/?f=2%5Ctheta%20%3D%20%5Cfrac%7B%5Cpi%7D%7B2%7D)
![\theta = \frac{\pi}{4}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Cfrac%7B%5Cpi%7D%7B4%7D)
Now, it is required to demonstrate that critical point leads to a maximum. The second derivative is:
![\frac{d^{2}x}{d\theta^{2}} = -2700\cdot \sin 2\theta](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5E%7B2%7Dx%7D%7Bd%5Ctheta%5E%7B2%7D%7D%20%3D%20-2700%5Ccdot%20%5Csin%202%5Ctheta)
![\frac{d^{2}x}{d\theta^{2}} = -2700](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5E%7B2%7Dx%7D%7Bd%5Ctheta%5E%7B2%7D%7D%20%3D%20-2700)
Which demonstrates the existence of the maximum associated with the critical point found before.
c) The equation for the vertical component of position is:
![y = 45\cdot t - 6\cdot t^{2}](https://tex.z-dn.net/?f=y%20%3D%2045%5Ccdot%20t%20-%206%5Ccdot%20t%5E%7B2%7D)
The maximum height can be found by deriving the previous expression, which is equalized to zero and critical values are found afterwards:
![\frac{dy}{dt} = 45 - 12\cdot t](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdt%7D%20%3D%2045%20-%2012%5Ccdot%20t)
![45-12\cdot t = 0](https://tex.z-dn.net/?f=45-12%5Ccdot%20t%20%3D%200)
![t = \frac{45}{12}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B45%7D%7B12%7D)
![t = 3.75\,s](https://tex.z-dn.net/?f=t%20%3D%203.75%5C%2Cs)
Now, the second derivative is used to check if such solution leads to a maximum:
![\frac{d^{2}y}{dt^{2}} = -12](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5E%7B2%7Dy%7D%7Bdt%5E%7B2%7D%7D%20%3D%20-12)
Which demonstrates the assumption.
The maximum height reached by the cannonball is:
![y_{max} = 45\cdot (3.75\,s)-6\cdot (3.75\,s)^{2}](https://tex.z-dn.net/?f=y_%7Bmax%7D%20%3D%2045%5Ccdot%20%283.75%5C%2Cs%29-6%5Ccdot%20%283.75%5C%2Cs%29%5E%7B2%7D)
![y_{max} = 84.375\,m](https://tex.z-dn.net/?f=y_%7Bmax%7D%20%3D%2084.375%5C%2Cm)