What Goes Up (a) It is well-known that the model in which air resistance is i...

What Goes Up

(a) It is well-known that the model in which air resistance is ignored, part (a) of Problem 36, predicts that the time t_a it takes the cannonball to attain its maximum height is the same as the time t_d it takes the cannonball to fall from the maximum height to the ground. Moreover, the magnitude of the impact velocity v_i 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 linear air resistance into account, compare the value of t_a with t_d and the value of the magnitude of v_i with v₀. A root-finding application of a CAS (or graphic calculator) may be useful here.

Reference:

Problem 36:

How High?—No Air Resistance Suppose a small cannonball weighing 16 lb is shot vertically upward 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 d ²s/dt ² = −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.

Problem 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 drag coefficient is k = 0.0025. [Hint: Slightly modify the DE in Problem 35.]