The maxima of a differential equation can be obtained by getting the 1st derivate dx/dy and equating it to 0.
Given the equation h = - 2 t^2 + 12 t , taking the 1st derivative result in:
dh = - 4 t dt + 12 dt
dh / dt = 0 = - 4 t + 12 calculating for t:
t = -12 / - 4
t = 3 s
Therefore the maximum height obtained is calculated by plugging in the value of t in the given equation.
h = -2 (3)^2 + 12 (3)
h = 18 m
This problem can also be solved graphically by plotting t (x-axis) against h (y-axis). Then assigning values to t and calculate for h and plot it in the graph to see the point in which the peak is obtained. Therefore the answer to this is:
The ball reaches a maximum height of 18 meters. The maximum of h(t) can be found both graphically or algebraically, and lies at (3,18). The x-coordinate, 3, is the time in seconds it takes the ball to reach maximum height, and the y-coordinate, 18, is the max height in meters.