Equation of velocity with quadratic drag, vertical throw

[Hannibal123]

Homework Statement

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

 

[haruspex]

Try http://hyperphysics.phy-astr.gsu.edu/hbase/airfri3.html

 

[Hannibal123]

Try http://hyperphysics.phy-astr.gsu.edu/hbase/airfri3.html

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
which leads to this
-√(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.

 

[haruspex]

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
which leads to this
-√(m/gk)⋅arctan(√(k/mg)⋅v)+c=t
however I'm not sure if this is correct, 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.

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

 

[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?

 

[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

 

[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)

 

[Hannibal123]

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

 

[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)})}