3. The ball's height above the field at the point it hits the stadium is "h" and "h=50 feet". The horizontal distance the ball travels is "s" and "s=400 feet". The angle of impact is "A" and "A=135 E". The height where the bat struck the ball is "k" and "k=3 feet". The distance that the ball went is "d".

Path of a Baseball Pitch

Parabola Project

Sean M.

Sean - Sections 9-7 and 12-2 should be helpful to you in finding the equation for your parabola. You may also find an equation online. You might also use Problem #13 on page 727 to find your parabola's equation. Finding the typical initial velocity of a baseball pitch and the height from which it's thrown will help. Good luck!

1. My parabola is going to be the path of a baseball pitch.

2. Projectile motion follows the flight path of a parabola, a curve generated by a point moving such that its distance from a fixed point on one axis is equal to its distance from a fixed line on the other axis. In other words, there is a proportional relationship between x and y throughout the trajectory or path of a projectile in motion. Most often this parabola can be visualized as a simple up-and-down curve like the shape of a domed roof.

4. In standard form, the equation would look like "h=ax^2+bx+k". I got the equation in standard form by figuring out that "c=k", since when "x=0", "y=k" and when "x=s", "y=h". In vertex form, the equation would look like "f(x)=a(x-h)^2+k". I got the equation easily because i knew that the vertex form was "f(x)=a(x-h)^2+k" and i just changed "x" to equal "s". Since the path of a baseball pitch is not a parabola you should plug in the numbers from step 3 into the equations "a=[s(tanA)-h+k]/s^2" and "b=[2h-s(tanA)-2k]/s", the path of the ball is in standard form with "a"=-.00279, "b"=1.235, and "c"=3. The home run distance, "d", is the positive solution to the equation, "(-.00279)x^2+(1.235)x+3=0". By using the quadratic formula we see that "d" is about 445 feet. From the data that we got, the parabolic path of the ball is "y=ax^2+bx", where "a"=-.00239 and "b"=1.199. So, the home run distance,"d", is the positive solution of "(-.00239)x^2+(1.199)x=0" and "d" is about 502 feet.

5. From now on we are going to use the same points, (0,3), (400,50), and (502,0) that we used for the other graph, but were converted into a quadratic equation using Excel and got this graph.

Standard form: y=-0.0012x^2+0.6017x+3
Vertex form: y-78.43=-.0012(x-250.71)^2
Factored form: We can't put it in factored form.

The vertex of the graph is (250.71, 78.43). It is a maximum because the graph is opening down. The vertex represents the greatest height that the ball could have gone.
The y-intercept of the graph is (0,3). The y-intercept represents the height of the bat at the time that it struck the ball.
The discriminate of the equation is .37644289, which means that there are two x-intercepts. The two x-intercepts are (-4.93,0) and (506.35,0).

## 3. The ball's height above the field at the point it hits the stadium is "h" and "h=50 feet". The horizontal distance the ball travels is "s" and "s=400 feet". The angle of impact is "A" and "A=135 E". The height where the bat struck the ball is "k" and "k=3 feet". The distance that the ball went is "d".

## Path of a Baseball Pitch

## Parabola Project

## Sean M.

Sean - Sections 9-7 and 12-2 should be helpful to you in finding the equation for your parabola. You may also find an equation online. You might also use Problem #13 on page 727 to find your parabola's equation. Finding the typical initial velocity of a baseball pitch and the height from which it's thrown will help. Good luck!

1. My parabola is going to be the path of a baseball pitch.

2. Projectile motion follows the flight path of a parabola, a curve generated by a point moving such that its distance from a fixed point on one axis is equal to its distance from a fixed line on the other axis. In other words, there is a proportional relationship between x and y throughout the trajectory or path of a projectile in motion. Most often this parabola can be visualized as a simple up-and-down curve like the shape of a domed roof.

4. In standard form, the equation would look like "h=ax^2+bx+k". I got the equation in standard form by figuring out that "c=k", since when "x=0", "y=k" and when "x=s", "y=h". In vertex form, the equation would look like "f(x)=a(x-h)^2+k". I got the equation easily because i knew that the vertex form was "f(x)=a(x-h)^2+k" and i just changed "x" to equal "s". Since the path of a baseball pitch is not a parabola you should plug in the numbers from step 3 into the equations "a=[s(tanA)-h+k]/s^2" and "b=[2h-s(tanA)-2k]/s", the path of the ball is in standard form with "a"=-.00279, "b"=1.235, and "c"=3. The home run distance, "d", is the positive solution to the equation, "(-.00279)x^2+(1.235)x+3=0". By using the quadratic formula we see that "d" is about 445 feet. From the data that we got, the parabolic path of the ball is "y=ax^2+bx", where "a"=-.00239 and "b"=1.199. So, the home run distance,"d", is the positive solution of "(-.00239)x^2+(1.199)x=0" and "d" is about 502 feet.

5. From now on we are going to use the same points, (0,3), (400,50), and (502,0) that we used for the other graph, but were converted into a quadratic equation using Excel and got this graph.

Standard form: y=-0.0012x^2+0.6017x+3

Vertex form: y-78.43=-.0012(x-250.71)^2

Factored form: We can't put it in factored form.

The vertex of the graph is (250.71, 78.43). It is a maximum because the graph is opening down. The vertex represents the greatest height that the ball could have gone.

The y-intercept of the graph is (0,3). The y-intercept represents the height of the bat at the time that it struck the ball.

The discriminate of the equation is .37644289, which means that there are two x-intercepts. The two x-intercepts are (-4.93,0) and (506.35,0).