Physics The equation h = -16t^2 + 32t + 9 gives the height of a ball, h, in feet above the ground, at t seconds
after the ball is thrown upward. How many seconds after the ball is thrown will it reach its maximum height? What is its maximum height?
Find the equation of the parabola that passes through each set of points.
To find the maximum height of the ball, we need to determine the vertex of the parabolic equation h = -16t^2 + 32t + 9. The formula for the x-coordinate of the vertex of a quadratic equation in the form ax^2 + bx + c is given by x = -b/2a.
In this case, a = -16 and b = 32, so the x-coordinate of the vertex is x = -32 / (2 * -16) = -32 / -32 = 1.
To find the corresponding y-coordinate of the vertex, we substitute the x-value back into the equation:
h = -16(1)^2 + 32(1) + 9
h = -16 + 32 + 9
h = 25
Therefore, the maximum height of the ball occurs 1 second after it is thrown, and its maximum height is 25 feet.
To find the equation of a parabola that passes through given points, we need three points because a parabola is defined by a quadratic equation. Let's assume the three points are (x1, y1), (x2, y2), and (x3, y3).
The general equation of a quadratic function is y = ax^2 + bx + c. By substituting each point into this equation, we can determine the values of a, b, and c.
The three points (x1, y1), (x2, y2), and (x3, y3) will give us three sets of equations:
For (x1, y1): y1 = a(x1)^2 + b(x1) + c
For (x2, y2): y2 = a(x2)^2 + b(x2) + c
For (x3, y3): y3 = a(x3)^2 + b(x3) + c
By solving these three equations simultaneously, we can find the values of a, b, and c, and substitute them into the general equation y = ax^2 + bx + c to find the equation of the parabola.