Q 29 - Precalculus Enhanced with Graphing Utilities

Question

According to physicist Peter Brancazio, the key to a successful foul shot in basketball lies in the

arc of the shot. Brancazio determined the optimal angle of the arc from the free-throw line to be 45 degrees. The arc also depends on the velocity with which the ball is shot. If a player shoots a foul shot, releasing the ball at a 45-degree angle from a position 6 feet above the floor, then the path of

the ball can be modeled by the function

$h\left(x\right)=-\frac{44x^2}{v^2}+x+6$

where h is the height of the ball above the floor, x is the forward distance of the ball in front of the foul line, and v is the initial velocity with which the ball is shot in feet per second. Suppose a player shoots a ball with an initial velocity of 28 feet per second.

(a) Determine the height of the ball after it has traveled 8 feet in front of the foul line.

(b) Determine h(12). What does this value represent?

(c) Find additional points and graph the path of the basketball.

(d) The center of the hoop is 10 feet above the floor and 15 feet in front of the foul line. Will the ball go through the hoop? Why or why not? If not, with what initial velocity must the ball be shot in order for the ball to go through the hoop?

Step-by-Step Solution

Answer

(a) The height of the ball after it has traveled 8 feet in front of the foul line is 10.4 ft

(b) $h\left(12\right)\approx 9.918 ft$

(c) The additional points are $\left(0,6\right), \left(6,9.98\right), \left(14,9\right) \left(20,3.55\right)$

(d) The  initial velocity must the ball be shot in order for the ball to go through the hoop is 30 ft/sec

Step 1. Given Information

According to physicist Peter Brancazio, the key to a successful foul shot in basketball lies in the

arc of the shot. Brancazio determined the optimal angle of the arc from the free-throw line to be 45 degrees. The arc also depends on the velocity with which the ball is shot. If a player shoots a foul shot, releasing the ball at a 45-degree angle from a position 6 feet above the floor, then the path of the ball can be modeled by the function

$h\left(x\right)=-\frac{44x^2}{v^2}+x+6$

where h is the height of the ball above the floor, x is the forward distance of the ball in front of the foul line, and v is the initial velocity with which the ball is shot in feet per second. Suppose a player shoots a ball with an initial velocity of 28 feet per second.

Part (a) of Step 1. The height of the ball

$h\left(x\right)=-\frac{44x^2}{v^2}+x+6$

Given initial velocity 28 feet per second. $v=28 ft/s$

Ball has travelled 8 feet. Let $x=8 ft$

Substitute 28 for v and 8 for x into the function h(x)

$$\begin{gathered} h\left(8\right)=-\frac{44{\left(8\right)}^2}{{\left(28\right)}^2}+8+6 \\ h\left(8\right)=10.4 ft \end{gathered}$$

Part (b) of Step 1. The height of the ball

$h\left(x\right)=-\frac{44x^2}{v^2}+x+6$

Given initial velocity 28 feet per second. $v=28 ft/s$

Ball has travelled 12 feet. Let $x=12$

Substitute 28 for v and 12 for x into the function h(x)

$$\begin{gathered} h\left(12\right)=-\frac{44(12{)}^2}{(28{)}^2}+\left(12\right)+6 \\ h\left(12\right)\approx 9.918 ft. \end{gathered}$$

Part (c) of Step 1. The path of the basketball.

The additional points can be found by substituting some values for x, in the given equation 

$h\left(x\right)=-\frac{44x^2}{v^2}+x+6$

Substitute $x=0$

$$\begin{gathered} h\left(0\right)=\frac{-44·0^2}{{28}^2}+0+6 \\ h\left(0\right)=6 \end{gathered}$$

Substitute $x=6$

$$\begin{gathered} h\left(6\right)=\frac{-44·6^2}{{28}^2}+6+6 \\ h\left(6\right)=9.98 \end{gathered}$$

Substitute $x=14$

$$\begin{gathered} h\left(14\right)=\frac{-44·{14}^2}{784}+14+6 \\ h\left(14\right)=9 \end{gathered}$$

Substitute $x=20$

$$\begin{gathered} h\left(20\right)=\frac{-44·{20}^2}{784}+20+6 \\ h\left(20\right)=3.55 \end{gathered}$$

The graph of the path of the basketball is

Part (d) of Step 1.

$h\left(15\right)=-\frac{44(15{)}^2}{{28}^2}+\left(15\right)+6\approx 8.4\text{ feet }$

When the ball is 15 feet in front of the (visible) foul line, it'll be below the hoop. Therefore, it can't go through the hoop. So, in order for the ball to pass through the hoop, we need to have $h\left(15\right)=10$.

$$\begin{gathered} 10=-\frac{44(15{)}^2}{v^2}+\left(15\right)+6 \\ {v}^2=4\left(225\right) \\ v=30 ft/sec \end{gathered}$$