# Equation of velocity with quadratic drag, vertical throw

## 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
$$v(0)=v_0$$

Related Introductory Physics Homework Help News on Phys.org
haruspex
Homework Helper
Gold Member

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.

haruspex
Homework Helper
Gold Member
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 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)?

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?

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
Homework Helper
Gold Member
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)

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

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