# Vertical motion with retarding force

1. Jan 26, 2008

### darkfall13

Hello everyone, I've used these forums quite a bit and found it very helpful thanks for all you kind souls helping us through the sciences, but today is my first post :P

I am deriving the velocity of a projectile fired vertically through a retarding force and I continue getting every step the same as the book until the last equation, it may be due to it being nearly 2 in the morning but I wanted to see others thoughts on it to help me learn the reasoning better. (And oh yeah this retarding force is only linear to the velocity)

so if we have:

-mg - km$$\dot{y}$$ = m$$\ddot{y}$$

we can easily work to

-g - k$$\dot{y}$$ = $$\ddot{y}$$

$$\frac{dv}{dt}$$ = -g - kv

dv = dt(-g - kv)

$$\int\frac{dv}{g + kv}$$ = -$$\int{dt}$$

$$\frac{1}{k}$$ ln(g + kv) = -t + c

ln(g + kv) = -kt + c

g + kv = $$e^{-kt + c}$$

This is where the book and I agree to

but then it arrives to

v = $$\frac{dy}{dt}$$ = -$$\frac{g}{k}$$ + $$\frac{kv_0 + g}{k} e^{-kt}$$

Can someone explain to me how it arrives there? Thank you so much!

2. Jan 26, 2008

### psi^

:)

Just notice, you can find c when t = 0, v = v0:

ln(g + kv) = -kt + c =>
ln(g + kv0) = -k.0 + c <=>
c = ln(g + kv0)

So,
g + kv = $$e^{-kt + c}$$ <=>
g + kv = $$e^{-kt + ln(g + kv0)}$$ <=>
g + kv = $$e^{-kt}$$.(g + kv0)

3. Jan 26, 2008

### darkfall13

ah ok perfect thanks a bunch I kept going around in circles with it last night.