Respuesta :
Answer:
kick 1 has travelled 15 + 15 = 30 yards before hitting the ground
so kick 2 travels 25 + 25 = 50 yards before hitting the ground
first kick reached 8 yards and 2nd kick reached 20 yards
Explanation:
1st kick travelled 15 yards to reach maximum height of 8 yards
so, it has travelled 15 + 15 = 30 yards before hitting the ground
2nd kick is given by the equation
y (x) = -0.032x(x - 50)
[tex]Y = 1.6 X - 0.032x^2[/tex]
we know that maximum height occurs is given as
[tex]x = -\frac{b}{2a}[/tex]
[tex]y =- \frac{1.6}{2(-0.032)} = 25[/tex]
and maximum height is
[tex]y = 1.6\times 25 - 0.032\times 25^2[/tex]
y = 20
so kick 2 travels 25 + 25 = 50 yards before hitting the ground
first kick reached 8 yards and 2nd kick reached 20 yards
Answer:
The second ball traveled more distance, horizontally and vertically.
Explanation:
The given function is
[tex]f(x)=-0.032x(x-50)[/tex]
To find the distances that the second football travels, we need to find the vertex of its movement, because it's movement has a parabola form. The quadratic expression is
[tex]f(x)=-0.032x^{2} +1.6x[/tex]
Where [tex]a=-0.032[/tex] and [tex]b=1.6[/tex]
The vertex has coordinates of [tex](h,k)[/tex], where
[tex]h=-\frac{b}{2a}[/tex]
Replacing values, we have
[tex]h=-\frac{1.6}{2(-0.032)}=25[/tex]
Then, [tex]k=f(h)[/tex]
[tex]k=f(25)=-0.032(25)^{2} +1.6(25)\\k=-20+40=20[/tex]
Which means the maximum height of the second football is 20 yards. That means it travels 40 yards vertically.
Now, its horizontal distance can be found when [tex]f(x)=0[/tex]
[tex]0=-0.032x^{2} +1.6x\\0=x(-0.032x+1.6)\\x_{1}=0\\ -0.032x_{2} +16=0\\x_{2}=\frac{-16}{-0.032}\\ x_{2}=500[/tex]
So, its horizontal distance is 500 yards.
Comparing the distances between the footballs.
Ball 1
Horizontal distance of 30 yards.
Vertical distance of 30 yards.
Ball 2
Horizontal distance of 500 yards.
Vertical distance of 40 yards.
If we find their difference, it would be
Horizontal: 500 - 30 = 470 yards.
Vertical: 40 - 30 = 10 yards.
Therefore, the second ball traveled more distance, horizontally and vertically.
