Function for a free throw trajectory of ball

Question

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

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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!

Answer 1

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.

Answer 2

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.