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- (a) It is well known that the model in which air resistance is ignored, part (a) of Problem 36, predicts that the time ta it takes the cannonball to attain its maximum height is the same as the time td it takes the cannonball to fall from the maximum height to the ground. Moreover, the magnitude of the impact velocity vi will be the same as the initial velocity v0 of the cannonball. Verify both of these results.
- (b) Then, using the model in Problem 37 that takes air resistance into account, compare the value of ta with td and the value of the magnitude of vi with v0. A root-finding application of a CAS (or graphic calculator) may be useful here.
36. How High?—No Air Resistance Suppose a small cannonball weighing 16 pounds is shot vertically upward, as shown in Figure 3.1.13, with an initial velocity v0 = 300 ft/s. The answer to the question “How high does the cannonball go?” depends on whether we take air resistance into account.
- (a) Suppose air resistance is ignored. If 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. Find the velocity v(t) of the cannonball at time t.
- (b) Use the result obtained in part (a) to determine the height s(t) of the cannonball measured from ground level. Find the maximum height attained by the cannonball.
FIGURE 3.1.13 Find the maximum height of the cannonball in Problem 36
37. How High?—Linear Air Resistance Repeat Problem 36, but this time assume that air resistance is proportional to instantaneous velocity. It stands to reason that the maximum height attained by the cannonball must be less than that in part (b) of Problem 36. Show this by supposing that the constant of proportionality is k = 0.0025. [Hint: Slightly modify the differential equation in Problem 35.]

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Chapter 3 Solutions
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