# Differential Eqn: Find an expression of the velocity of a rock.

1. Sep 17, 2014

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

A 20 kg rock falls with the force of gravity acting on it. Another force acts on the rock (same direction as gravity) that is proportional to the velocity squared, 100v^2. Come up with an expression of the rock's velocity at any time t.

2. Relevant equations

Hint: (d/dx)(tan^-1(x)) = 1/(1 + x^2)

3. The attempt at a solution

F(net) = ma = Fg + 100v^2

ma = mg + 100v^2
a = g +(100v^2)/m (m=20)
dv/dt = g + 5v^2
dv/dt = g((5/g)v^2 + 1)
dv/((5/g)v^2 + 1) = g(dt)
dv/((root(5/g)*v)^2 +1) = g(dt) (integrate both sides)
tan^-1(root(5/g)*v) = gt + C
root(5/g)*v = tan(gt + C)

v(t) = (root(g/5))tan(gt + C)

I want to know if I can get rid of the constant as well as if my approach to the general solution is correct. Also if my general soln is correct.

Thanks!

2. Sep 17, 2014

### Staff: Mentor

You cannot ignore units like that. 20 and 20kg are completely different things.
The force given as 100v^2 has mismatching units, are you sure there is nothing else given there?

You cannot get rid of the constant as you do not know the initial velocity.

3. Sep 17, 2014

That's all that the question gives. There's no initial conditions given either. I assumed that the additional force would equal 100v^2 as it says it's proportional. But you're right, the units don't match up.. Could it be possible that the units are hidden within the coefficient of v^2?

4. Sep 17, 2014

### olivermsun

Yes, there has to be a dimensional coefficient with the proper units.

Normally the force proportional to v2 would be something like the drag force, which acts in the direction opposite to the velocity, hence against gravity in the example. But oh well.