# Equation of velocity with quadratic drag, vertical throw

1. Dec 16, 2012

### Hannibal123

1. The problem statement, all variables and given/known data

I am having trouble deriving the velocity equation for an objekt moving vertically upwards. the net force will be
$$m*dv/dt=-mg-kv^2$$
where k is the drag. At the time t=0 the object's velocity wil be
$$v(0)=v_0$$

2. Dec 16, 2012

### haruspex

3. Dec 16, 2012

### Hannibal123

im not sure how they get to that equation, and why they are using y, which seems to be an height.
I have tried seperating the variables, and have come to this integral
$$-√(m/gk)∫1/(1+u^2 )du=∫dt$$
where u is a substitution defined by
$$u=√(k/mg)⋅v$$
$$-√(m/gk)⋅arctan(√(k/mg)⋅v)+c=t$$
however I'm not sure if this is corrrect, and the constant c must be solved for the condition t=0 and v(0)=v_0. Furthermore this will have to be solved for v, because i need the velocity equation.

4. Dec 16, 2012

### haruspex

That all looks correct. Can you not rearrange it easily enough into the form v = f(t)?

5. Dec 17, 2012

### Hannibal123

for the start condition (t,v)=(0,v_0) i can only cook c down to this
$$c=√(m/gk)*arctan(√(k/mg)*v_0 )$$
which is going to give a pretty nasty equation when inserted into the previous equation, that i want to solve for v?

6. Dec 17, 2012

### Hannibal123

which i eventually solve to this for v
$$tan((t-√(m/gk)⋅arctan(√(k/mg)⋅v_0 ))/√(m/gk))/√(k/mg)$$

or in a more pretty picture: http://imgur.com/V53iV

but I'm not sure if it's correct, or if it can be shortened

7. Dec 17, 2012

### haruspex

arctan(v) = t+c
v = tan(t+c) = (tan(t)+tan(c))/(1- tan(t)tan(c))
v0 = tan(c)
v = (tan(t)+v0)/(1- tan(t)v0)

8. Dec 17, 2012

### Hannibal123

I don't really follow you there, would you mind elaborating that?

9. Dec 17, 2012

### haruspex

$$-√(m/gk)⋅arctan(√(k/mg)⋅v)+c=t$$
$$arctan(√(k/mg)⋅v)=\frac{c-t}{√(m/gk)}$$
$$√(k/mg)⋅v=tan(\frac{c-t}{√(m/gk)})$$
$$=tan(\frac{c}{√(m/gk)}-\frac{t}{√(m/gk)})$$
$$=\frac{tan(\frac{c}{√(m/gk)})-tan(\frac{t}{√(m/gk)})}{1+tan(\frac{c}{√(m/gk)})tan(\frac{t}{√(m/gk)})}$$
At t=0:
$$√(k/mg)⋅v_0=tan(\frac{c}{√(m/gk)})$$
$$√(k/mg)⋅v=\frac{√(k/mg)⋅v_0-tan(\frac{t}{√(m/gk)})}{1+√(k/mg)⋅v_0 tan(\frac{t}{√(m/gk)})}$$
$$v=\frac{v_0-\frac{1}{√(m/gk)}tan(\frac{t}{√(m/gk)})}{1+√(k/mg)⋅v_0 tan(\frac{t}{√(m/gk)})}$$