Question

a ball is thrown upward. its height ( h, in feet ) is modeled by the function h = -16t^2 + 64t+3, where t is the length of time ( in seconds ) that the ball has been in the air. what is the maximum height the ball reaches? a ball is thrown upward. its height ( h, in feet ) is modeled by the function h = -16t^2 + 64t+3, where t is the length of time ( in seconds ) that the ball has been in the air. what is the maximum height the ball reaches?

Answer 1

The maximum height reached by the ball is 67 feet, found by determining the vertex of the quadratic equation which represents the ball's trajectory.

Step-by-step explanation:

The question involves finding the maximum height reached by a ball thrown upward, described by the quadratic equation h = -16t^2 + 64t + 3. To find the maximum height, we need to determine the vertex of the parabola represented by this equation, as the coefficient of t^2 is negative, indicating that the parabola opens downward. The vertex form of a quadratic equation is given by h = a(t-h)^2 + k, where (h, k) is the vertex of the parabola. The time at which the maximum height occurs can be found by using the formula t = -b/(2a), where a and b are the coefficients of t^2 and t, respectively, from the original equation.

In this case, a = -16 and b = 64, so the time of maximum height is t = -64/(2(-16)) = 2 seconds. Substituting t = 2 back into the original equation gives us the maximum height: h = -16(2)^2 + 64(2) + 3. After calculation, the maximum height is found to be h = -64 + 128 + 3 = 67 feet.

Answer 2

To find the maximum height just simply find the vertex of  -16t^2 + 64t + 3 and to find the axis of symmetry or the x value of the vertex do -b/2a or -64/-32 = 2 in this situation. Plug in to get the y value -16(2)^2 + 64(2) + 3 = 67 Vertex (2,67) So the max height is 67 feet (takes 2 seconds to do so)