A body of mass 6 kg is projected vertically upward with an initial velocity 15 m

Question

A body of mass 6 kg is projected vertically upward with an initial velocity 15 meters per second.

The gravitational constant is g = 9.8 m/s2. The air resistance is equal to k|v| where k is a constant.  

Find a formula for the velocity at any time ( in terms of k ):
v(t) = ?

Find the limit of this velocity for a fixed time t0 as the air resistance coefficient k goes to 0. ( Enter t0 as tzero . )
v(t0) = ?

How does this compare with the solution to the equation for velocity when there is no air resistance?

This illustrates an important fact, related to the fundamental theorem of ODE and called 'continuous dependence' on parameters and initial conditions. What this means is that, for a fixed time, changing the initial conditions slightly, or changing the parameters slightly, only slightly changes the value at time t.  

The fact that the terminal time t under consideration is a fixed, finite number is important. If you consider 'infinite' t, or the 'final' result you may get very different answers. Consider for example a solution to y = y, whose initial condition is essentially zero, but which might vary a bit positive or negative. If the initial condition is positive the "final" result is plus infinity, but if the initial condition is negative the final condition is negative infinity.

Differential Equations | 3.1 | Zill

Explanation / Answer

The net force acting on the object is F = -mg - kv.

Gravity always acts downward so that's why the sign is negative. The minus sign in front of the kv term comes from the fact that air resistance always acts in the opposite direction of motion. Using Newton's 2nd Law, get the differential equation

F = -mg - kv = ma = mv'

with initial condition v(0) = 15.

Now we need to solve the initial value problem via an integrating factor.

Divide through by m and move terms around to get

v' + (k/m)v = -g

To which the solution is

v(t) = -mg/k + C*e^(-kt/m).

Applying the initial condition gives

v(0) = -mg/k + C = 15 ==> C = 15+ mg/k = 15 + 6*9.8/k = 15 + 58.8/k.

So the solution is

v(t) = -58.8/k + (15 + 58.8/k)*e^(-kt/m) = 15e^(-kt/m) + (58.8/k)(e^(kt/m)-1).

In the limit as k ->0, the first term goes to 15 and the second term goes to 58.8*t/m so

v(t0) = 15 + 58.8*t/m.

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