# How to integrate the following and solve for $v$: $\frac{dv} {((kv^2)/m - g)} = dt$?

I am trying to find the effect of air resistance on the projectile motion of an ice skater performing a jump (I am aware it technically isn't called a projectile motion once there is air resistance involved but for the sake of explaining it shortly, I will say it is).

I have found that $$F\text{net} = m*a = kv^2 - mg$$ , since gravity is acting downwards and air resistance acting upwards, k being some constant for air resistance. So for the math part:

$$\frac{dv} {((kv^2)/m - g)} = dt$$

I have done integration by substitution at first, by using u = kv^2/m - g and $$du = (\frac{2kv}m)dv$$ , however, I am not sure if this is the correct approach and even if it is, I am unable to solve the final equation I found for $$v$$:

$$\ln\left(\frac{kv^2}m - g\right)= \frac{t2kv}m$$

steps I followed to integrate and solve for v

I essentially followed the steps I outlined in the attached image

If anyone can help I would appreciate it so much. Thank you!

• If the ice skater is moving upward (i.e. when $v>0$) then the force due to air resistance and the force of gravity both point downward. On the other hand, if the ice skater is moving downward (i.e. when $v<0$), then the force of gravity still points downward while air resistance points upward. So you actually have that $F_{\text{net}}=-mg-kv|v|$ assuming, of course, that the magnitude of air resistance is directly proportional to the square of the object's velocity.
– user801306
Commented Feb 21, 2022 at 21:23
• If the equation is correct, one can rewrite as $$a v^2-g-e^{2 t a v}=0, a=\frac km$$ which may not be solvable using the W Lambert function Commented Feb 21, 2022 at 21:35

Simply use this Riccati's equation as a special type of first-order nonlinear differential equation and the solution $$v(t)=-\frac{\sqrt{b} \tanh\big(\sqrt{ab} (c_1+t)\big)}{\sqrt{a}}$$ where

• $$a=\frac{k}{m}$$,
• $$b=\frac{m g}{k}$$ and
• $$c_1$$ is the integration constant.

Mind that arctanh has a special relationship to the logarithm.

Some straightforward rewrite and the results are as expected. $$arctanh$$ is the inverse of $$tanh$$.

Somewhat comfortable is this with g, k and m.

To do the solution is documented in done by wolframaplha too but in the paid for version.

There are of course sides that show the solution. For example ode0123. That is a specialized side for ordinary differential equations.