A body of mass 3 kg is projected vertically upward with an i

A body of mass 3 kg is projected vertically upward with an initial velocity 73 meters per second. The gravitational constant is g = 9.8m/s?. 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 ): Find the limit of this velocity for a fixed time t_0 as the air resistance coefficient k goes to 0. (Write t_0 as t0 in your answer.) 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.

Solution

a.) Here, the ball will undergo a combination of two forces causing its deceleration.

First is the gravitational pull = 3g and the second is air resistance = KV

Now, the net deceleration at any point would be g+kv/3

That is dv/dt = -(g+kv/3)

That is, dv = -(g+kv/3) dt which can be written as dv/-(g+kv/3) = dt

We integrate both the sides, with initial time being 0 and initial velocity being 73 m/s

That is, (-3/k) [ln(g+kv/3) - ln(g+ 73k/3)] = t

we rearrange, ln(g+kv/3) = ln(g+73k/3) - kt/3

That is, g+kv/3 = (g+73k/3) x e^- kt/3

V = 3/k[((g+73k/3) x e^- kt/3) - g] = [((3g+73k) x e^- kt/3) - 3g]/k

b.) Now, for k tending to zero, the above velocity will be limiting towards 0/0 form, hence we differentiate the numerator and denominator [L\'Hospital rule]

We get V(limk->0) = [73 x e^- kt/3] - [(t/3) x (3g+73k) x e^- kt/3]

Now we find the limit for k tending to zero, we obtain the right hand side as, 73 - gt

This equation is same as that for no resistance (V = U- GT)

NOTE: We you are not aware of the L\'Hostpital\'s rule, I will quickly tell you how to go about it. If the limiting value of an expression tends to be of the form of infinity by infinity or zero/zero, then we differentiate the numerator and denominator separately and apply the limiting again.

What we basically do there is that we determine the \'rate of change\' of the values of denominator and numerator, either towards zero or infinity. The one which approaches the final value quickly governs the actual limit.