An object is thrown straight up into the air at an initial vertical velocity of 32 feet per second from an initial height of 128 feet and is represented by the vertical motion model, h(t) = -16t^2 + 32t + 128. How much time does it take for the object to reach its maximum height?
as with any quadratic, the vertex will be at t = -b/2a
To find the time it takes for the object to reach its maximum height, we need to find the vertex of the parabolic function represented by the equation h(t) = -16t^2 + 32t + 128.
The vertex of a parabola in the form y = ax^2 + bx + c is given by the formula: t = -b/2a.
Comparing the equation h(t) = -16t^2 + 32t + 128 to the standard form y = ax^2 + bx + c, we have:
a = -16
b = 32
Using the formula, we can calculate the time as follows:
t = -b/2a
t = -32 / (2 * -16)
t = -32 / -32
t = 1
Therefore, it takes 1 second for the object to reach its maximum height.
To find the time it takes for the object to reach its maximum height, we need to determine the vertex of the parabolic function h(t) = -16t^2 + 32t + 128. The vertex represents the maximum point of the function.
The formula to find the t-coordinate of the vertex of a parabola in the form ax^2 + bx + c is given by t = -b / (2a).
In our case, a = -16, b = 32, and c = 128. Plugging these values into the formula, we get:
t = -32 / (2 * -16)
t = -32 / -32
t = 1
Therefore, it takes 1 second for the object to reach its maximum height.