A baseball player hits a foul ball into the stands. He wants to know where the ball would've landed if the stands had not been there. The points they give you are: The y-intercept (0,1), (2,2.5) and (134,18). I'm pretty sure you are looking for the zeroes.
assuming the ball's trajectory to be a parabola,
y = ax^2 + bx + c
Plug in the three points and solve for y=0, because that;'s how high the ball would have been when it hit the ground.
1 = a*0 + b*0 + c
so, c=1
y = ax^2 + bx + 1
2.5 = 4a + 2b
18 = 17956a + 134b
multiply the first by 67 and subtract from the second, and we have
a = -0.00845
b = 1.2669
s, we have
y = -.00845x^2 + 1.2669x + 1
When y=0, x = 150.714
Must have hit into the stands about halfway to the outfield fence.
To find where the foul ball would have landed if the stands were not there, we can use the information provided to find the equation of the trajectory of the ball.
Given that we have three points: (0,1), (2,2.5), and (134,18), we can assume that the trajectory of the ball can be represented by a quadratic equation in the form of y = ax^2 + bx + c.
To find the equation of the trajectory, we can start by substituting the coordinates of the given points into the equation.
When we substitute the coordinates (0,1) into the equation, we get:
1 = a(0)^2 + b(0) + c
1 = 0 + 0 + c
c = 1
So, we have found that c = 1.
Next, let's substitute the coordinates (2,2.5) into the equation:
2.5 = a(2)^2 + b(2) + 1
2.5 = 4a + 2b + 1
4a + 2b = 1.5 ...(1)
Now, substitute the coordinates (134,18) into the equation:
18 = a(134)^2 + b(134) + 1
18 = 17956a + 134b + 1
17956a + 134b = 17 ...(2)
We now have a system of two equations with two variables (a and b) to solve simultaneously. Using these equations, we can determine the values of a and b, which will allow us to find the equation of the trajectory.
Solving equations (1) and (2) simultaneously can be done through various methods, such as substitution or elimination. For simplicity, let's use the substitution method.
From equation (1), we can express a in terms of b:
4a + 2b = 1.5
a = (1.5 - 2b) / 4
Substituting this expression for a into equation (2), we get:
17956[(1.5 - 2b) / 4] + 134b = 17
(4489 - 4491.5b) + 134b = 17
-4491.5b + 134b = 17 - 4489
-4357.5b = -4472
b = -4472 / -4357.5
b ≈ 1.027
Now, substitute this value of b back into equation (1) to find a:
4a + 2(1.027) = 1.5
4a + 2.054 = 1.5
4a ≈ 1.5 - 2.054
4a ≈ -0.554
a ≈ -0.1385
Now we have determined the values of a and b as approximately -0.1385 and 1.027, respectively.
So, the equation of the trajectory is:
y = -0.1385x^2 + 1.027x + 1
To find where the ball would have landed if the stands were not there, we need to find the x-coordinate at which y = 0 (where the ball hits the ground).
Setting y = 0 in the equation gives us:
0 = -0.1385x^2 + 1.027x + 1
To solve this quadratic equation, we can use various methods such as factoring, completing the square, or using the quadratic formula. In this case, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values of a, b, and c from the equation gives us:
x = (-(1.027) ± √((1.027)^2 - 4(-0.1385)(1))) / (2(-0.1385))
Simplifying this equation will allow us to find the x-coordinates:
x = (-1.027 ± √(1.0549 + 0.554)) / (-0.277)
x = (-1.027 ± √(1.6089)) / (-0.277)
x = (-1.027 ± 1.269) / (-0.277)
Using the plus-minus symbol, we get two possible values for x:
x₁ = (-1.027 + 1.269) / (-0.277) ≈ 0.882
x₂ = (-1.027 - 1.269) / (-0.277) ≈ 8.499
Hence, if the stands were not there, the foul ball would have landed at approximately (0.882, 0) and (8.499, 0).