Function for a free throw trajectory of ball I've been trying to write functions for basketball's free throw trajectory. 
These are main points:


*

*the three throw line is 15 feet away from the foot of the basket

*the basket is 10ft above the ground

*the player releases the ball from approximate height of 8ft



I have to find two different functions for this information.
Let's denote the position of the basket as A, and the position of player as B, then:
$A(0;10)$ and $B(15;8)$.
First, the standard quadratic form:
$\ y = ax^{2} + bx + c$
Since 10 is the height required for the object to hit the basket -
$\ y = ax^{2} + bx + 10$
From my calculations, If c in function was greater or smaller than 10, Ball wouldn't hit the basket.
Then, considering that ball is shot from 8 feet and the distance from the starting point to the basket is 15 feet:
$\ 8 = a(15)^{2} + b(15) + 10$
Now to calculate the terms, I need to obtain the value of variable b, Thus, I get axis of symmetry, which is obviously 7.5.
Since:
$\ x[0] = 7.5$
$\frac{-b}{2a}=7.5$
$\ b=-15a$ 
Now substitution:
$\ 8 = 225a + (-15a)(15) + 10$
$\ 8 = 225a - 225a + 10$
$\ 8 = 10$
$\ 2$
I'm not sure afterwards, since i don't seem to properly obtain a, what could be the problem? Are the values that i have obtained correct?
So, what could be the proper function which would land the object in basket by it's graph?
Thanks!
 A: Use the SUVAT equations of motion:
$$v=u+at$$
$$s=\tfrac{1}{2}(u+v)t$$
$$s=ut+\tfrac{1}{2}at^2$$
$$s=vt-\tfrac{1}{2}at^2$$
$$v^2=u^2+2as$$
Where $s$ is displacement, $u$ is initial velocity, $v$ is final velocity, $a$ is acceleration and $t$ is time.
You will need to resolve horizontally and vertically, giving $s_x,u_x,v_x,a_x$ and $s_y,u_y,v_y,a_y$. 
Time is time.
A: There is one degree of freedom
Assume that the initial velocity of the throw is $p{\bf i} + q{\bf j}$, so $u_x=p$ and $u_y=q$. The acceleration due to gravity is $0{\bf i} - g{\bf j}$ , meaning that $a_x=0$ and $a_y=-g$. The throw point is 15ft from the basket, so $s_x=15$. The initial throw height is 8ft above the ground and the basket height is 10ft, so $s_y=2$.
This takes the throw point as the origin. 
Considering only vertical motion: $s=ut+\frac{1}{2}at^2$ gives $15 = qt-\frac{1}{2}gt^2$.
Considering only horizontal motion: $s=ut+\frac{1}{2}at^2$ gives $2 = pt$.
Solving $2=pt$ gives $t=\frac{2}{p}$, assuming that $p \neq 0$, and so $15=q\left(\frac{2}{p}\right)-\frac{1}{2}g\left(\frac{2}{p}\right)^2$
$$15p^2=2pq-2g$$
$$q=\frac{15p^2+2g}{2p}$$
For any choice of $p$, you get a $q$ which gives the initial velocity.
