Please explain in detail so I can understand how to do it:
A football player punts a ball. The path of the ball can be modeled by the equation y = -0.004x^2 + x +2.5, where x is the horizontal distance, in feet, the ball travels and y is the height in feet, of the ball. How far from the football player will the ball land? Round to the nearest tenth of a foot.
Here is a graph showing the path of the ball
http://www.wolframalpha.com/input/?i=0%3D+-0.004x%5E2+%2B+x+%2B2.5
So the x-intercept would be the distance covered
so -0.004x^2 + x +2.5 = 0
use the quadratice equation were
a = -.004, b = 1, and c = 2.5
x = (-1 ± √1.04)/-.008
= appr. 252.8 or a negative
To find the point where the ball will land, we need to determine the value of x when y equals zero. This is because when the ball lands, its height will be zero.
Given the equation y = -0.004x^2 + x + 2.5, we can set y to zero and solve for x. So, we have:
0 = -0.004x^2 + x + 2.5
To solve this quadratic equation, we can use the quadratic formula. The quadratic formula is:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = -0.004, b = 1, and c = 2.5. Substituting these values into the quadratic formula, we get:
x = (-1 ± √(1^2 - 4(-0.004)(2.5))) / (2(-0.004))
Simplifying further:
x = (-1 ± √(1 + 0.04)) / (-0.008)
x = (-1 ± √1.04) / (-0.008)
Now, we evaluate the two solutions separately:
x1 = (-1 + √1.04) / (-0.008)
x2 = (-1 - √1.04) / (-0.008)
Calculating the values:
x1 ≈ 92.52
x2 ≈ -603.52
Since x represents the distance traveled by the ball in feet, a negative value does not make sense in this context. Therefore, we can ignore x2.
Therefore, the ball will land approximately 92.52 feet from the football player. Rounded to the nearest tenth, the ball will land approximately 92.5 feet from the player.
To find how far from the football player the ball will land, we need to determine the x-coordinate when the ball reaches its maximum y-coordinate. This x-coordinate represents the horizontal distance traveled by the ball before it starts descending.
Step 1: Determine the vertex of the parabolic equation.
The equation representing the path of the ball is y = -0.004x^2 + x + 2.5. This is a quadratic equation in the form of y = ax^2 + bx + c, where a = -0.004, b = 1, and c = 2.5. The x-coordinate of the vertex (horizontal distance when the ball reaches its maximum height) can be found using the formula x = -b / (2a).
Plugging in the values, we get:
x = -(1) / (2 * -0.004)
x = 1 / 0.008
x = 125
Step 2: Determine the y-coordinate at the vertex.
Substituting the calculated x-coordinate (125) into the equation, we can find the maximum height (y-coordinate) reached by the ball.
y = -0.004(125)^2 + (125) + 2.5
y = -0.004(15625) + 125 + 2.5
y = -62.5 + 125 + 2.5
y = 65
Step 3: Determine the distance traveled by the ball.
The horizontal distance traveled by the ball depends on the time it takes to reach the maximum height and return to the ground. This can be found using the equation t = 2 * (maximum height / gravity), where gravity is approximately 32.2 ft/s^2.
t = 2 * (65 / 32.2)
t ≈ 4.036
Finally, we can calculate the distance traveled by multiplying the time by the horizontal speed of the ball.
Distance = speed * time
The speed of the ball can be determined by differentiating the equation y with respect to x and finding the derivative:
dy/dx = -0.008x + 1
Evaluating the derivative at the vertex (125) gives us the speed:
dy/dx = -0.008(125) + 1
dy/dx = -1 + 1
dy/dx = 0
This indicates that the speed of the ball at the vertex is zero, meaning the horizontal component of the ball's velocity is constant. Therefore, the distance traveled by the ball is simply the time multiplied by the initial horizontal velocity, which is given by the coefficient of x in the equation:
Distance = (0.008)(125)(4.036)
Distance ≈ 0.4036(125)
Distance ≈ 50.45 ft
Therefore, the ball will land approximately 50.45 feet from the football player.