A baseball player swings and hits a pop fly straight up in the air to the catcher. The height of the baseball in meters t seconds after it is hit is given by a quadratic function h(t) = −4.9t^2 + 9.8t + 1. What is the maximum height of the baseball?
If you know Calculus,
h ' (t) = -9.8t + 9.8 = 0 for a max of h
9.8t = 9.8
t = 1
h(1) = -4/9(1) + 9.8 + 1 = 5.9 m
If you don't know Calculus, complete the square
h(t) = -4.9 [t^2 - 2t +1 - 1 ] +1
= -4.9 [ (t-1)^2 - 1 ] + 1
= -4.9(t-1)^2 + 4.9 + 1
= -4.9(t-1)^2 + 5.9
so the vertex is (1,5.9) and the parabola opens downwards
So we have a maximum of 5.9 m when t = 1 sec
Bleh
To find the maximum height of the baseball, we need to determine the vertex of the quadratic function h(t) = -4.9t^2 + 9.8t + 1.
The x-coordinate of the vertex of a quadratic function in the form ax^2 + bx + c is given by -b/2a. In this case, a = -4.9 and b = 9.8.
Therefore, the x-coordinate of the vertex is -9.8 / (2 * -4.9) = -9.8 / -9.8 = 1.
To find the y-coordinate of the vertex, we substitute the x-coordinate (1) into the equation:
h(1) = -4.9(1)^2 + 9.8(1) + 1 = -4.9 + 9.8 + 1 = 6.9.
Therefore, the maximum height of the baseball is 6.9 meters.
To find the maximum height of the baseball, we need to determine the vertex of the quadratic function.
The vertex of a quadratic function in the form f(x) = ax^2 + bx + c can be found using the formula x = -b / (2a).
In this case, the quadratic function is h(t) = -4.9t^2 + 9.8t + 1.
Comparing this with the standard form, we have a = -4.9 and b = 9.8.
Using the formula x = -b / (2a), we can calculate the value for t as follows:
t = -9.8 / (2 * -4.9) = -9.8 / -9.8 = 1
So the maximum height of the baseball occurs at t = 1 second.
Now we can substitute t = 1 back into the quadratic function to find the maximum height:
h(1) = -4.9(1)^2 + 9.8(1) + 1
h(1) = -4.9 + 9.8 + 1
h(1) = 5.9
Therefore, the maximum height of the baseball is 5.9 meters.