# Linear motion problems solved with differential equations?

A body of mass m, which is initially at rest, falls vertically from the height h under the influence of constant gravitational acceleration, g = 9.81 ms−2 .

Assuming that the body’s air resistance is proportional to the square of its velocity v and opposite to the direction of v, i.e. F = −kv^2 , with k a positive constant, determine the velocity vimpact with which the body hits the ground level (height zero).

I have created this relation in the system:

$F=mg-kv^2=ma=m\dot{v}=mv*\frac{\mathrm{dv}&space;}{\mathrm{d}&space;x}$

However I am very stuck where to move from here, any guidance would be greatly appreciated.

Thanks

i think your equation is given by $$mg-k\left(\frac{dx}{dt}\right)^2=m\frac{d^2x}{dt^2}$$ substituting $$u(t)=\frac{dx(t)}{dt}$$ then we get $$\frac{m\frac{du(t)}{dt}}{gm-ku(t)^2}=1$$ integrating both sides we obtain $$\frac{\sqrt{m}arctanh\left(\frac{\sqrt{k}u(t)}{\sqrt{gm}}\right)}{\sqrt{gk}}=t+C_1$$

• Once I have derived this how can I obtain V-impact? – QuestionAnswer13 Mar 2 '18 at 15:33

I think you have no problem to derive $$g-Av^2=\frac{dv}{dt}$$ where $A=k/m$.

Now seperating variables: $$dt=\frac{dv}{g-Av^2}$$

Integrate both sides: $$t+C=-\frac{\ln\left(\frac{|Av-\sqrt{Ag}|}{|Av+\sqrt{Ag}|}\right)}{2\sqrt{Ag}}$$(by wolfy)

After some algebra I think you are able to make $v$ the subject.

$C$ is determined my initial conditions, which, in this case, I think is the initial velocity.