Suppose a small cannonball weighing 16 pounds is shot vertically upward, with an initial velocity v0 = 290 ft/s. The answer to the question "How high does the cannonball go?" depends on whether we take air resistance into account. If air resistance is ignored and the positive direction is upward, then a model for the state of the cannonball is given by d2s/dt2 = −g (equation (12) of Section 1.3). Since ds/dt = v(t) the last differential equation is the same as dv/dt = −g, where we take g = 32 ft/s2. If air resistance is incorporated into the model, it stands to reason that the maximum height attained by the cannonball must be less than if air resistance is ignored.

first one

step-by-step explanation:

a

step-by-step explanation:

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  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

so the change in y-value for function f is (6 )) = 9.

the table tells us

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).