Free-throw Shots

[nycmath]

Precalculus
Michael Sullivan
Edition 10
Chapter 2, Section 2.2

See attachments.

My setup for part (d) is as follows:

O = [-(44)(15)^2]/[(v)^2] + [15 + 6]

Is my setup correct to find the correct velocity?

Here is part (d):

The centerof the basket 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 ball?

Note:

h = height of ball above the floor

x = forward distance of the ball in front of the foul line

The problem is modeled by the function

h(x) = [-(44)x^2]/[v^2] + x + 6

[Math Tutor]

Your setup is almost right - the only problem is the left-hand side. You set the height equal to 0, but it should be 10.

The center of the hoop is 15 ft in front of the foul line and 10 ft above the floor, so the ball goes through when h(15) = 10. Setting h(15) = 0 would mean the ball is on the floor 15 ft out, which is not what we want.

First, check the given speed v = 20 ft/s:

[tex]h(15)=\frac{-44(15)^2}{20^2}+15+6=-24.75+21=-3.75[/tex]

The height is negative, so the ball has already hit the floor before it travels 15 ft. It does not go through the hoop.

Now solve for the velocity that works:

[tex]\frac{-44(15)^2}{v^2}+15+6=10[/tex]

[tex]\frac{-9900}{v^2}=-11[/tex]

[tex]v^2=900 \quad\Rightarrow\quad v=30[/tex]

We keep the positive root since v is a speed. So the ball must be shot with an initial velocity of 30 ft/s.

Quick check: [tex]h(15)=\frac{-9900}{900}+21=-11+21=10[/tex] - right at the rim.

[Eigenvalue]

Yes,this is mostly correct. However, the height should be set equal to 10

[nycmath]

Your setup is almost right - the only problem is the left-hand side. You set the height equal to 0, but it should be 10.

The center of the hoop is 15 ft in front of the foul line and 10 ft above the floor, so the ball goes through when h(15) = 10. Setting h(15) = 0 would mean the ball is on the floor 15 ft out, which is not what we want.

First, check the given speed v = 20 ft/s:

[tex]h(15)=\frac{-44(15)^2}{20^2}+15+6=-24.75+21=-3.75[/tex]

The height is negative, so the ball has already hit the floor before it travels 15 ft. It does not go through the hoop.

Now solve for the velocity that works:

[tex]\frac{-44(15)^2}{v^2}+15+6=10[/tex]

[tex]\frac{-9900}{v^2}=-11[/tex]

[tex]v^2=900 \quad\Rightarrow\quad v=30[/tex]

We keep the positive root since v is a speed. So the ball must be shot with an initial velocity of 30 ft/s.

Quick check: [tex]h(15)=\frac{-9900}{900}+21=-11+21=10[/tex] - right at the rim.

Thank you. I knew something was off with my setup. Thank you so much.

[nycmath]

Yes,this is correct.

According to Math Tutor my set up should be equal to 10 not 0. So, I was wrong.

[Eigenvalue]

Yes,this is correct.

According to Math Tutor my set up should be equal to 10 not 0. So, I was wrong.

I did not see the 10. the rest of your set up is correct

[nycmath]

Yes,this is correct.

According to Math Tutor my set up should be equal to 10 not 0. So, I was wrong.

I did not see the 10. the rest of your set up is correct

Thank you again for the quick reply.