Question

uppose 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. (a) Assume air resistance is proportional to instantaneous velocity. If the positive direction is upward, a model for the state of the cannonball is given by m dv dt = −mg − kv, where m is the mass of the cannonball and k > 0 is a constant of proportionality. Suppose k = 0.0025 and find the velocity v(t) of the cannonball at time t.

Answer 1

Given in the explanation

Explanation:

Given

w = 16 pounds

v₀ = 290 ft/s

g = 32 ft/s²

k = 0.0025 (Kg/s)

`m(dv)/(dt)= -mg - kv^(2)`

Solving the differential equation we obtain

`v(t)=((1)/(0.0125))*tan((-2*(t+C_(1) )/(5) )`

If v(0) = 290 ft/s, we have

`290=((1)/(0.0125))*tan((-2*(0+C_(1) )/(5) )`

⇒ C₁ = -3.254

Finally, we have

`v(t)=((1)/(0.0125))*tan((-2*(t-3.254 )/(5) )`