Q 30 - Precalculus Enhanced with Graphing Utilities

Question

The last player in the NBA to use an underhand foul shot (a “granny” shot) was Hall of Fame

forward Rick Barry who retired in 1980. Barry believes that current NBA players could increase their free-throw percentage if they were to use an underhand shot. Since underhand shots are released from a lower position, the angle of the shot must be increased. If a player shoots an underhand foul shot, releasing the ball at a 70-degree angle from a position 3.5 feet above the floor, then the path of the ball can be modeled by the function h(x) = -

$h\left(x\right)=-\frac{136x^2}{v^2}+2.7x+3.5$

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.

(a) The center of the hoop is 10 feet above the floor and 15 feet in front of the foul line. Determine the initial velocity with which the ball must be shot in order for the ball to go through the hoop.

(b) Write the function for the path of the ball using the velocity found in part (a).

(c) Determine h(9). What does this value represent?

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

Step-by-Step Solution

Answer

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

(b) The function for the path of the ball using the velocity found in part (a) is $h\left(x\right)=-\frac{136x^2}{{30}^2}+2.7x+3.5$

(c) $h\left(9\right)=15.56$

(d) The graph of the path of the basketball is 

Step 1. Given information

The last player in the NBA to use an underhand foul shot (a “granny” shot) was Hall of Fame

forward Rick Barry who retired in 1980. Barry believes that current NBA players could increase their free-throw percentage if they were to use an underhand shot. Since underhand shots are released from a lower position, the angle of the shot must be increased. If a player shoots an underhand foul shot, releasing the ball at a 70-degree angle from a position 3.5 feet above the floor, then the path of the ball can be modeled by the function h(x) = -

$h\left(x\right)=-\frac{136x^2}{v^2}+2.7x+3.5$

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.

Part (a) of Step 1. The initial velocity of the ball

Given equation $h\left(x\right)=-\frac{136x^2}{v^2}+2.7x+3.5$

The center of the hoop is 10 feet above the floor, then let $h\left(x\right)=10$

There is 15 feet in front of the foul line, then let $x=15$

Substitute 15 for x and 10 for h(x) into the function

$$\begin{gathered} h\left(x\right)=-\frac{136x^2}{v^2}+2.7x+3.5 \\ 10=-\frac{136(15{)}^2}{v^2}+2.7\left(15\right)+3.5 \\ 10=-\frac{30600}{v^2}+44 \\ {v}^2=900 \\ v=30 ft/sec \end{gathered}$$

Part (b) of Step 1. The function for the path of the ball

Substitute the value of $v=30 ft/sec$ in the given equation

$h\left(x\right)=-\frac{136x^2}{{30}^2}+2.7x+3.5$

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

To find the height of the ball after it has traveled 9 feet in front of the foul line substitute $x=9$ in the given equation

$$\begin{gathered} h\left(x\right)=-\frac{136x^2}{900}+2.7x+3.5 \\ h\left(9\right)=-\frac{136·9^2}{900}+2.7·9+3.5 \\ h\left(9\right)=15.56 \end{gathered}$$

Part (d) of Step 1. Graph of the path

One point on the graph is (9,15.66). Let's find some more. Chose some values for x.

Substitute $x=3$ in the given function

$$\begin{gathered} h\left(3\right)=\frac{-136·3^2}{900}+2.7·3+3.5 \\ h\left(3\right)=10.24 \end{gathered}$$

Substitute $x=5$ in the given function

$$\begin{gathered} h\left(5\right)=\frac{-136·5^2}{900}+2.7·5+3.5 \\ h\left(5\right)=13.22 \end{gathered}$$

Substitute $x=13$ in the given function

$$\begin{gathered} h\left(13\right)=\frac{-136·{13}^2}{900}+2.7·13+3.5 \\ h\left(13\right)=13.06 \end{gathered}$$

The plot of the points are