0
$\begingroup$

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:

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

Thanks

$\endgroup$
1
$\begingroup$

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 $$\frac{\sqrt{m}arctanh\left(\frac{\sqrt{k}u(t)}{\sqrt{gm}}\right)}{\sqrt{gk}}=t+C_1$$

$\endgroup$
  • $\begingroup$ Once I have derived this how can I obtain V-impact? $\endgroup$ – QuestionAnswer13 Mar 2 '18 at 15:33
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.