Freefall with terminal velocity - expressions for velocity and position

Question

A body weighing 29 kg is dropped from a height of 30 m with an initial velocity of 3m/sec. Assume that the air resistance is proportional to the velocity of the body. if the limiting velocity is known to be 39 m/sec, find:

1. an expression for the velocity of the body at any time t
2. an expression for the position of the body at any time t

Seems to be out of this world for me, as this is supposed to be done with differential equations. Any help is much appreciated!

Answer 1

Start from Newton's 2nd law. I will write time derivatives as dots. $$m\ddot{x} = mg - b\dot{x}$$ Where b is some constant of proportionality. Simplify this down to $$\ddot{x} = g-\frac{b}{m}\dot{x}$$ This is just a first order differential in $\dot{x}$ In other words: $$\frac{d}{dt}\dot{x}+\frac{b}{m}\dot{x}=g$$ Get an integrating factor $$I(t) = e^{\int\frac{b}{m}dt} = e^\frac{bt}{m}$$ This reduces our differential equation to $$\frac{d}{dt}\left[\dot{x}e^\frac{bt}{m}\right] = e^\frac{bt}{m}g$$ Integrating will give you your expression for velocity. Once you have velocity you can integrating again to get displacement. You can get all the unknowns such as $b$ and the integration constants from your boundary conditions.

Answer 2

If the velocity is $v$ then

$$ \dot{v} = g - \beta v $$

This is directly integrated with

• Time as a function of velocity (to be inverted) $$t = \int \frac{1}{\dot{v}}\,\mathrm{d}v = \int \limits_{v_1}^v \frac{1}{g-\beta v}\,\mathrm{d}v = - \frac{1}{\beta} \ln \left( \frac{g-\beta v}{g-\beta v_1}\right)$$
• Distance as a function of velocity $$ x-x_1 = \int \frac{v}{\dot{v}}\,\mathrm{d}v = \int \limits_{v_1}^v \frac{v}{g-\beta v}\,\mathrm{d}v =\frac{v_1-v}{\beta}-\frac{g}{\beta^2} \ln \left(\frac{g-\beta v}{g-\beta v_1}\right)$$
• Subsitute $v(t)$ into the above $x(v)$ to get $$x -x_1= \frac{g-\beta v_1}{\beta^2} \mathrm{e}^{-\beta t} + \frac{\beta ( g t+v_1)-g}{\beta^2} $$