Why does solving $\int \frac{v}{9.8-0.0025v^2}\mathrm{d}v=\int1{d}x$ for $v^2$ in terms of $x$ produce 2 completely different answers?

Question

In this question $g=9.8$ (acceleration of free fall). You are also given that when $x=0$ $v=0$.

My answer is $v^2=400g(1-e^\frac{x}{200})$. I obtained it by integrating both sides so that $x=-200\ln(g-0.0025v^2) + A$ [where $A$ is the constant of integration]. Substituting boundary conditions to obtain $A$ yields $A =200lng$.

This is how the other answer was obtained (uploaded it from the mark-scheme of the exam paper):

Its clear what the problem here is: it depends which side of the equation you choose to place the constant of integration. How can this be?

By the way here is the exam question:

I'd really appreciate it if someone could explain what it is i'm doing wrong.

Thank you.

Answer 1

I think it's just an algebra error. Starting from your expression, $$x=-200\ln(g-0.0025v^2) + A=200\ln\frac{g}{g-0.0025v^2},$$ and exponentiating both sides, you get $$e^{0.005x}=\frac{g}{g-0.0025v^2}.$$ Now multiply both sides by $e^{-.005x}(g-.0025v^2)$ and solve for $v^2.$

Answer 2

So to begin with since the particle is falling vertically speed here is just $|v|$

Now on to finding $v$. The most standard method to do so is solve for it as a function of time and then convert to function of height.

Function of time: $$ m \frac{d^2 x}{dt^2} = \frac{1}{400} \left( \frac{dx}{dt} \right)^2 - mg$$

We have air resistance counterbalanced by gravity for net force. Solving this is quite simple, merely observe that we can create a variable $u = \frac{dx}{dt}$ to convert the equation to:

$$ m \frac{du}{dt} = \frac{1}{400}u^2 - mg$$

Yielding

$$ \frac{m}{\frac{1}{400}u^2 - mg} du = dt$$ $$ u = 20\sqrt{mg}\tanh \left(\frac{1}{2} \sqrt{\frac{g}{m}}t + C\right)$$ Integrate via trigonometric integrals and solve for $u$, now integrate AGAIN to solve for $x$

$$ x = \frac{20\sqrt{mg}}{\frac{1}{2} \sqrt{\frac{g}{m}}} \log \left( \cosh\left( \frac{1}{2} \sqrt{\frac{g}{m}}t + C\right) \right) $$

Do what you gotta do from there (substituting values etc)

Answer 3

Below, the correction of the sign mistake in your answer. By the way, avoid to use the same symbol $A$ for two different constants $A$ and $B$