body of mass 6 kg is projected vertically upward with an initial velocity 39 met

Question

body of mass 6 kg is projected vertically upward with an initial velocity 39 meters per second (thus a negative velocity).

The gravitational constant is g=9.8m/s2 (NOT 9.81). The air resistance is equal to k|v| where k is a constant. Remember the equation mv?=mg?kv (friction for velocity upwards is increasing the downward speed, so -k v is still correct for upward motion)

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

Find the terminal velocity: v=

Find the limit (might need L'Hospital's rule) of this velocity for a fixed time t0 as the air resistance coefficient k goes to 0. (Write t0 as t0 in your answer.)
v(t0)=

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 Lae, we get the differential equation

F = -mg - kv = ma = mv'

with initial condition v(0) = 39. 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 = 39 ==> C = 39 + mg/k = 39 + 6*9.8/k =39+ 58.8/k.

So the solution is

v(t) = -58.8/k + (39 + 58.8/k)*e^(-kt/m) = 39e^(-kt/m) + (58.8/k)(e^(kt/m)-1).

In the limit as k ->0, the first term goes to 39 and the second term goes to 58.8*t/m so

v(tt) = 39 + 58.8*t/m.

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