Make a table of values, plot and connect the obtained points.
B
b
The distance the ball falls from t=0 to t=1 can be found by subtracting the h-values from these two points.
C
c
One way to solve this is to compare the changes in the h-values on the graph.
A
a
B
b
16 feet
C
c
No, see solution.
Practice makes perfect
a
We will graph h=-16t^2+72. Our independent variable is t, so it will be on the horizontal axis. The dependent variable is h, so the vertical axis will be labeled h. Note that negative values of time t and height h do not make sense in this problem, so we will only graph positive values of t and h.
We will now make a table of points for us to graph.
t
-16t^2+72
h=-16t^2+72
0
-16( 0)^2+72
72
0.5
-16( 0.5)^2+72
68
1
-16( 1)^2+72
56
1.5
-16( 1.5)^2+72
36
2
-16( 2)^2+72
8
Now we plot and connect these points.
b
To find how far the ball has fallen from t=0 to t=1, we need to subtract the h value at t=0 from the h value at t=1. We already computed the h values in the table above.
t
-16t^2+72
h=-16t^2+72
0
-16( 0)^2+72
72
1
-16( 1)^2+72
56
We will subtract 56 from 72 to find how far the ball has fallen.
72-56=16ft
We can see that between t=0 and t=1, the ball fell 16 feet.
c
The distance that the ball has fallen between two points can be found from the h values. The distance the ball fell between t=0 and t=1 can be represented by the vertical segment below.
Now, let's compare this to the segment that represents how far the ball fell from t=1 to t=2.
We can see that the segment representing t=1 to t=2 is much larger. This represents a larger change in the h-values. Therefore, the ball falls a greater distance from t=1 to t=2.
