<span>The maxima of a differential equation can be obtained by
getting the 1st derivate dx/dy and equating it to 0.</span>
<span>Given the equation h = - 2 t^2 + 12 t , taking the 1st derivative
result in:</span>
dh = - 4 t dt + 12 dt
<span>dh / dt = 0 = - 4 t + 12 calculating
for t:</span>
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:
<span>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.</span>
Since they tell us that this is linear, having a constant rate of change, we can express this as a line:
y=mx+b, where m=slope (change in y divided by change in x) and b=y-intercept (value of y when x=0)
First find the slope, or m, which mathematically is:
m=(y2-y1)/(x2-x1), in this case:
m=(880-440)/(2000-1000)
m=440/1000
m=0.44, so far our line is:
y=0.44x+b, now we can use either data point to solve for b, I'll use (1000,440)
440=0.44(1000)+b
440=440+b
0=b, so our line is just:
y=0.44x
The answer to the problem is c
