S0lve for $v(t)$ when there is a quadratic drag force

Question

How do I get $v(t)$ in this problem?

The motion of a falling body is governed by Newton's second law $\frac{d(mv)}{dt} = mg- F_d.$

Find $v(t)$ for $m=1kg$, $g = 4 \frac{m}{s^2}$ and $F_d = v^2$.

Answer 1

We have $$v' = g - v^2$$ (the m's are omitted because they equal 1 kg) There are various ways to solve this, one of which is to identify this as a separable equation (i.e. the kind of differential equation you learn to compute in a first or second semester calculus course). In general, when we have a differential equation of the form $$y' = f(x)h(y)$$ its solutions satisfy $$\int \frac{1}{h(y)} dy = \int f(x) dx$$ Here, we have $f(t) = 1$ and $h(v) = 4 - v^2$. Therefore, $$\frac{\ln(\frac{v+2}{v-2})}{4} = \int \frac{1}{4 - v^2} dv = \int 1 dt = t + C$$ (using partial fractions for the left hand integral)

Solving for $v$, we obtain $$v = \frac{2ce^{4t} - 2}{ce^{4t} + 1}$$ The value of $c$ depends on your initial conditions.

Answer 2

Note first that by linearity of derivative, we have $$ \frac{d(mv)}{dt} = m\frac{dv}{dt} $$

So now we rewrite the equation: $$m \frac{dv}{dt} = mg - v^2$$ $$\implies \frac{dv}{dt} = g - \frac{1}{m}v^2 $$

$$\implies \int\frac{dv}{1- \frac{1}{mg}v^2} = g\int dt$$

Now we use the change of variables $$z^2 = \frac{1}{mg}v^2 \implies z dz = \frac{1}{mg} v dv \implies dv = \sqrt{mg} dz $$

This implies that $$\sqrt{mg} \int \frac{dz}{1-z^2} = gt + C \implies \sqrt{mg}\ arctanh(z) = gt +C $$

$$\implies arctanh\frac{v}{\sqrt{mg}} = \frac{gt}{\sqrt{mg}} + C \implies v(t) = \sqrt{mg} \tanh \left( \frac{gt}{\sqrt{mg}} \right) + C'$$

The value of $C'$ depends on the initial conditions which weren't given.

See for example here: http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/quadrag.html#c2 or http://www.physics.udel.edu/~szalewic/teach/419/cm08ln_quad-drag.pdf