Finding x-coordinate of a parabola's vertex given the y-coordinates of two points and the distance between them

Question

I'm given the following problem:

1. Suppose a basketball player has shot a basketball, whose trajectory describes a parabola, and the player scores said shot. The ball was 2 meters above the floor when it left the player's hand and the hoop was 5 meters ahead, and 3 meters above the floor. It is known that the highest point of the basketball's trajectory was 6 meters above the floor. What was the horizontal distance between the player and the point where the ball reached it's highest point?

$$f(x) = -\alpha {(x - h)^\beta }$$

Let's suppose, for the sake of convenience, that $\beta = 2, \alpha=1$ and let's place the vertex at the origin of the graph. Then: $$f(x) = -{x^2}$$

$$\eqalign{ & {\text{if a = - 2, then }}f(a) = - 4 \cr & {\text{and if b = }}\sqrt 3 {\text{, then }}f(b) = - 3 \cr & {\text{Thus:}} \cr & - a + b = 2 + \sqrt 3 = 5 \cr} $$

This doesn't make any sense? I must find a $k$ such that $|c-a|=k$ when $f(c)=0$.

Answer 1

Don't know why you put $\alpha=1$. If I understood correctly your problem is $$ \begin{gathered} \left\{ \begin{gathered} y = - \alpha x^2 \hfill \\ y\left( a \right) = - \alpha a^2 = - 4 \hfill \\ y\left( {a + 5} \right) = - \alpha \left( {a + 5} \right)^2 = - 3 \hfill \\ \end{gathered} \right. \hfill \\ \frac{4} {{a^2 }}\left( {a + 5} \right)^2 = 3 \hfill \\ a^2 + 40a + 100 = 0 \hfill \\ a = \frac{{ - 40 \pm \sqrt {1600 - 400} }} {2} = - 20 \pm 10\sqrt 3 \hfill \\ \left\{ \begin{gathered} a = - 20 + 10\sqrt 3 \cong - 2.68\quad \alpha = 4/a^2 \cong 0.557 \hfill \\ a = - 20 - 10\sqrt 3 \cong - 37.32\quad \alpha = 4/a^2 \cong 0.00287 \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} $$ where the second solution corresponds to the basket being centered from below (with a rifle probably).

Answer 2

This may not be the approach that the authors of the problem had in mind, but here’s how I’d go about solving this.

Assume no air resistance and, as is usual at this scale, that gravitational acceleration is constant, so the path of the ball is parabolic. It’s launched with some initial velocity $(v_x,v_y)$. The vertical velocity becomes zero at the peak, so by conservation of energy, $\frac12mv_y^2=4mg$, or $v_y=\sqrt{8g}$. The time it takes to reach the peak is $t_1=v_y/g=\sqrt{8g}/g=\sqrt{8/g}$. The time to fall back down into the basket is found by solving $\frac12gt_2^2=3$ for $t_2$, which gives $t_2=\sqrt{6/g}$.

The horizontal velocity is constant, so the horizontal distances are in the same proportions as the travel times. Therefore, the horizontal distance from the shooter to the peak is $${t_1\over t_1+t_2}\cdot5={\sqrt{8/g}\over\sqrt{8/g}+\sqrt{6/g}}\cdot5=20-10\sqrt3\approx2.68m.$$ This is a bit over half of the distance, which we’d expect from the basket’s being higher than the shooter.