1)When Twinkle Toes Tony kicked a football, it landed 100ft from where he kicked it. It also reached a height of 125 ft. Write an equation that, when graphed, will model the path of the ball from the moment it was kicked until it first touched the ground.
2) A new skateboarder's ramp just arrived at Bungey's Family Fun Center. A cross -sectional view shows that the shape is parabolic. The sides are 12 ft high and 15 ft apart. Write an equation that, when graphed, will show the cross section of this ramp.
1) Oh, Twinkle Toes Tony sure knows how to kick a ball! Let's model the path of the ball with an equation. We can use the standard equation of a parabola: y = ax^2 + bx + c. Since we know the ball landed 100 ft away and reached a height of 125 ft, we can assume the vertex of the parabola is at (100, 125). This gives us the equation y = a(x - 100)^2 + 125.
2) Ah, a skateboard ramp at Bungey's Family Fun Center! Let's visualize the cross section of this ramp with a parabolic shape. With sides that are 12 ft high and 15 ft apart, we can assume the vertex of the parabola is at (0, 12). This gives us the equation y = ax^2 + 12.
Now, let's hope those skateboarders don't get too carried away and start doing acrobatics like Twinkle Toes Tony!
1) To model the path of the ball, we can use a quadratic equation in the form of:
y = ax^2 + bx + c
where:
- y represents the height of the ball
- x represents the horizontal distance from where the ball was kicked
Since the ball reached a height of 125 ft, we know that when x = 0 (the moment it was kicked), y = 125. Therefore, the equation becomes:
y = ax^2 + bx + 125
To find the coefficients a, b, and c, we need more information. Let's use the fact that the ball landed 100 ft from where it was kicked. This means that when the ball touches the ground, y = 0 and x = 100. Substituting these values into the equation, we get:
0 = a(100)^2 + b(100) + 125
Simplifying further:
10000a + 100b + 125 = 0
We still need one more piece of information to solve for a, b, and c.
2) The equation for the cross-sectional view of the ramp can be written in the form of:
y = ax^2 + bx + c
where:
- y represents the height of the ramp
- x represents the distance from the center of the ramp
The sides of the ramp are 12 ft high, which means that when x = ±7.5 (half of the distance between the sides), y = 12. Therefore, we have two points: (7.5, 12) and (-7.5, 12).
Using these values, we can set up a system of equations to find the coefficients a, b, and c:
12 = a(7.5)^2 + b(7.5) + c
12 = a(-7.5)^2 + b(-7.5) + c
This system of equations can be solved to determine the values of a, b, and c.
1) To model the path of the ball, we can use a quadratic equation since the shape of the path is parabolic. Let's assume that the origin is the initial position where the ball was kicked from.
Step 1: Determine the equation for the vertical motion of the ball.
The path of the ball can be modeled by a quadratic equation in the form of y = ax^2 + bx + c, where y is the height and x is the horizontal distance from the origin.
Since the ball reaches a maximum height of 125 ft, it means that the vertex of the parabola is at (0, 125).
So, the equation for the vertical motion becomes: y = a(x - 0)^2 + 125
Simplifying it further, it becomes: y = ax^2 + 125
Step 2: Determine the equation for the horizontal motion of the ball.
Since the ball landed 100 ft from where it was kicked, when it lands, the horizontal distance (x) would be 100 ft, and the vertical height (y) would be 0.
Substituting these values into the equation we derived from Step 1:
0 = a(100)^2 + 125
Now, solve for 'a':
10000a + 125 = 0
10000a = -125
a = -125 / 10000
a = -0.0125
Step 3: Combine the equations for both vertical and horizontal motion.
The equation for the path of the ball becomes:
y = -0.0125x^2 + 125
So, when graphed, this equation will model the path of the ball from the moment it was kicked until it first touched the ground.
2) Since the shape of the ramp is parabolic, we can use a quadratic equation to model its cross section.
Let's consider the origin (0, 0) as the bottom point of the parabolic ramp.
The height (y) of the ramp at any given horizontal distance (x) from the vertex can be represented by the quadratic equation y = ax^2, where a is a constant.
Given that the sides of the ramp are 12 ft high and 15 ft apart, we can use this information to find the value of 'a'.
At the point where the sides of the ramp are 12 ft high, the height (y) is 12 and the horizontal distance (x) is half the width of the ramp, which is 15/2 ft.
Substituting these values into the equation, we have:
12 = a(15/2)^2
To simplify calculation, let's find a common denominator:
12 = a(225/4)
Now, solve for 'a':
a = (12 * 4) / 225
a = 48 / 225
a ≈ 0.213
Therefore, the equation for the cross section of the ramp is:
y = 0.213x^2
When graphed, this equation will show the cross section of the ramp at Bungey's Family Fun Center.
Both problems require to graph the vertex.
1. we know that the parabola passes through the following points: (0,0), (50,125),(100,0).
Using the standard formula, where (h,k) represent the coordinates of the vertex, we have h=50, k=125, and the parabola is:
y=a(x-h)^2+k=a(x-50)^2+125
Since the parabola passes through (0,0), we have
0=a(0-h)^2+k = ah^2+k = 2500a+125
=> a=-125/2500=-1/20
So parabola is
y=-(x-50)^2/20+125
(2) can be solved similarly.