How to work out the trajectories of the cannons in Mario 64 So recently I've played one of my childhood games again, namely Super Mario 64, and as anyone who has played it as well knows, you will find cannons at specific locations that allow Mario to send himself flying in the direction he is pointing at, even allowing him to change its rotational angle. Now obviously the trajectory resembles a parabola, and obviously the parabola's general shape (as in its coefficient a) depends on the cannon's angle, and (correct me if I am wrong, physicists) on how much Mario gets accelerated by the cannon. Some quick research said 
Mario's mass = 90kg = m_mario
I couldn't really come up with a velocity through research, so I am just gonna assume that
v = 20m/s 
v0 = 0m/s
The acceleration from firing the cannon feels rather fast, so I am gonna assume
a = (20-v0)/ 0.3 = 66m/s².
Therefore, f_Cannon = 90*66 = 5940N (just in case that's gonna be of use later on)
Now my first question is: I'm sure there must be a way of sort of "converting" the rotational angle and f_cannon into a correlating coefficient "a" for the parabola trajectory. I suspect trigonometric functions might be helpful, but I can't really think of a way to solve this problem.
Question #2: I also wondered if it would be possible to find out what "a" would have to be equal to in oder for the parabola to have a specific root apart from x = 0 (assuming b = 1 and c = 0, always). 
Example: Say at x = 7 there is a power star, and Mario wants to collect the star hovering in the air just before he hits the ground at x = 7. if the parabola = ax²+x, then it is possible to find a: f(a) = 49a+7
a = (-7/49)
Here are my personal figures for solving this problem
How to find a for the root x = 7, assuming b = 1 and c = 0
Please excuse any mathematical errors of mine, I'm just your average freshman who likes math a lot, lol.
Thanks!
 A: Let's try to simplify this:


*

*Mario leaves the cannon at the origin $(0,0)$ with a speed $V_0$ at an angle $\theta$ measured from the horizontal, so has a horizontal component of velocity of $V_0 \cos(\theta)$ and an initial vertical component of velocity of $V_0 \sin(\theta)$

*Assuming no air resistance, at time $t$ Mario will have travelled $x=V_0 \cos(\theta) t$ horizontally, and due to gravity $g$ will be at a height $y = V_0 \sin(\theta) t - \frac12 g t^2$

*In terms of $x$ we have $t = \frac{1}{V_0 \cos(\theta)}x$, and substituting gives the parabola
$$y = - \frac{g}{2V_0^2\cos^2(\theta) }x^2  +\tan(\theta) x$$  
If you want this to also hit the $x$-axis at the particular point $(x_1,0)$ then that requires $\frac{g x_1}{V_0 } = 2 \sin(\theta) \cos(\theta) = \sin(2\theta) $ so $$\theta = \frac12\sin^{-1} \left( \frac{g x_1}{V_0^2 }  \right)$$ being aware that there can be more than one angle with a given sine
Taking your example numbers of $V_0=20$ and $x_1=7$ together with $g=9.8$ you would get $$\theta \approx 0.0862 \text{ or } 1.4846 \text{ radians}$$ i.e. about $\theta \approx 5^\circ \text{ or } 85^\circ$, either a very shallow trajectory or a very high one.  That would give the approximate parabolas $$y\approx -0.01234 x^2 + 0.08639x$$ 
$$y\approx -1.65363 x^2 +11.57542x$$ 
