Jose bautista hits a baseball that travels for 142m before it lands. The flight of the ball can be modelled by a quadratic function in which x is the Horizontal distance the ball has travelled away from jose, and h(x) is the height Vertical distance of the ball at that distance.
Assume that the ball was between 0.6m and 1.5m above the ground when it was hit.
1. determine an equation that models the path of the ball, given this additional info:
-the ball was 1.2 m off the ground when it was hit.
-the ball reached a max of 17m in height at approximately 70m away from Jose.
Explain the method you are using for this equation.
So I know that the vertex is (70, 17).
But I'm lost on how to do this equation.
Please help!!
College of education Kangere
Mathematics & intersciece
To determine an equation that models the path of the ball, we can use the vertex form of a quadratic equation:
h(x) = a(x - h)^2 + k
where (h, k) represents the vertex of the parabola and "a" represents the coefficient that determines the shape of the parabola.
Given that the vertex is (70, 17), we can substitute these values into the equation to get:
h(x) = a(x - 70)^2 + 17
To find the value of "a", we can utilize the information that the ball was 1.2m off the ground when it was hit. Substitute this into the equation:
1.2 = a(0 - 70)^2 + 17
Simplifying this equation, we get:
1.2 = 4900a + 17
Subtracting 17 from both sides:
1.2 - 17 = 4900a
-15.8 = 4900a
Dividing both sides by 4900:
a = -15.8 / 4900
a ≈ -0.003224
Hence, the equation that models the path of the ball is:
h(x) = -0.003224(x - 70)^2 + 17
This equation represents a quadratic function that describes the vertical distance of the ball at a horizontal distance x from Jose.
To model the path of the ball, we can start by assuming that the quadratic function representing the height of the ball with respect to its horizontal distance is of the form:
h(x) = ax^2 + bx + c
We are given several key points of information:
1. The ball was 1.2 m off the ground when it was hit, which gives us the point (0, 1.2).
2. The ball reached a maximum height of 17 m at approximately 70 m away from Jose, which gives us the vertex (70, 17).
Using these points, we can substitute them into the quadratic function and create a system of equations to solve for the unknown constants a, b, and c.
From the point (0, 1.2):
h(0) = a(0)^2 + b(0) + c = 1.2
c = 1.2
From the vertex (70, 17):
h(70) = a(70)^2 + b(70) + 1.2 = 17
Simplifying this equation, we get:
4900a + 70b + 1.2 = 17
Now, we have two equations:
1. a + b + 1.2 = 1.2
2. 4900a + 70b + 1.2 = 17
Solving this system of equations will give us the values of a and b. Substituting these values back into the equation h(x) = ax^2 + bx + c will give us the final equation that models the path of the ball.
Your vertex is indeed (70,17)
Other points would be (142,0) and (0,1.2)
Since the vertex must lie midway between the two intercepts and the distance between 70 and 142 is 72, the other intercept must be (-2,0)
h(x) = a(x-70)^2 + 17 must be the equation
use (142,0) to find a
0 = a(72)^2 + 17
a = -17/5184 = appr -.00327932
h(x) = -.00327932(x-70)^2 + 17
check: does (0,1.2) lie on it?
LS = 1.2
RS = -.00327932(-70)^2 + 17 = appr .93
Were were expecting 1.2 , which is between .9 and 1.5 as originally stated.
But then it also said: " the ball was 1.2 m off the ground when it was hit. " , which appears to contradict the results of my equation.
If we take (0, 1.2) as fact , let's use that to find a
1.2 = a(-70)^2 + 17
a = -15.8/4900 = appr -.0032245
in that case, if x = 142, we should get zero
h(0) = -.0032245(72)^2 + 17 = .28 , close but mathematically too great an error.
It appears to me that some of the data is slightly off.