If a baseball is projected upward from ground level with an initial velocity of 64 feet per second, then its height is a function of time given y s(t) = -16t^2 + 64t. Graph this function for 0 < t<4, what is the maximum height?
Tv = -B/2A = -64 / -32 = 2 s.
Y = -16*2^2 + 64*2 = 64 Ft, = Max. ht.
V(T,Y) = (2,64).
Complete the chart below for graphing.
(T,Y).
(0.5,28)
(1.0,48).
(1.5,
V(2.0,64).
(2.5,
(3.0,
(3.5,
To graph the function y = -16t^2 + 64t, you can follow these steps:
1. Choose a set of time values, t, within the given range of 0 < t < 4. For example, you can choose t = 0, 0.5, 1, 1.5, 2, 2.5, 3, 3.5, and 4.
2. Plug each time value into the equation y = -16t^2 + 64t to find the corresponding height value, y.
For example, when t = 0, y = -16(0)^2 + 64(0) = 0. So the first point is (0, 0).
3. Repeat step 2 for all the time values you chose, and record the coordinates (t, y) for each point.
4. Plot the points on a coordinate plane.
5. Connect the points using a smooth, continuous, upward-curving line. This represents the graph of the function y = -16t^2 + 64t.
Now, to find the maximum height of the baseball, we need to determine the vertex of the parabolic function y = -16t^2 + 64t. The vertex represents the highest point on the graph.
The equation of the vertex can be found using the formula t = -b / (2a), where a and b are the coefficients of the quadratic equation. In this case, a = -16 and b = 64.
So, t = -64 / (2 * (-16)) = -64 / (-32) = 2.
Therefore, the time at which the maximum height occurs is t = 2. Plug t = 2 back into the equation y = -16t^2 + 64t to find the corresponding y value.
y = -16(2)^2 + 64(2) = -64 + 128 = 64.
Hence, the maximum height of the baseball is 64 feet.