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Assuming every parameter is constant except for the variable $t$, how does the author get from

enter image description here

to here enter image description here ?

When I integrate velocity myself I get $$z(t)=v_0t+\frac{v_e(m_0-qt)(\ln(1-\frac{qt}{m_0})-1)}{q} -\frac{gt^2}{2}$$ which leads to totally different results when using the same parametric and variable values. Specifically I can't understand how does he get rid of the $-1$ in the second parenthesis, and where does $(v_0+v_e)t$ come from.

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3 Answers 3

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There is an implicit assumption that you may have overlooked here.

In physics, the convention is that the initial position is zero, i.e. $z(0) = 0$. Therefore, you should first write your integral as,

$$z(t) = Z + v_0t+\frac{v_e(m_0-qt)}{q} \left[ \ln\left(1-\frac{qt}{m_0}\right)-1 \right]-\frac{gt^2}{2} \tag{1}$$

Then, find out $Z$ by setting $z(0)=0$, which yields,

$$Z = \frac{v_e m_0}{q}$$

Now, plug the value of $Z$ back into (1), you get

$$z(t) = (v_0+v_e)t+\frac{v_e(m_0-qt)}{q}\ln \left(1-\frac{qt}{m_0} \right) -\frac{gt^2}{2} $$

The resulting $z(t)$ is guaranteed to be zero at $t = 0 $.

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  • $\begingroup$ I forgot to take the integration constant into account, mainly because I was thinking at it as a definite integral $\int_0^t'v(t)dt$. Thank you. :) $\endgroup$
    – Albert
    Aug 17, 2019 at 8:53
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Your calculation

\begin{align}\frac{v_e\Big(m_0-qt\Big)\Big(\ln(1-\frac{qt}{m_0})-1\Big)}{q}&=v_e\Big(\frac{m_0}{q}-t\Big)\Big(\ln\big(1-\frac{qt}{m_0}\big)-1\Big)\\&=v_e\Big(\frac{m_0}{q}-t\Big)\Big(\ln\big(1-\frac{qt}{m_0}\big)\Big)-v_e\Big(\frac{m_0}{q}-t\Big)\\&=v_e\Big(\frac{m_0}{q}-t\Big)\Big(\ln\big(1-\frac{qt}{m_0}\big)\Big)+v_et-v_e\Big(\frac{m_0}{q}\Big)\end{align}

differs from $z(t)$ by the constant value of $v_e\Big(\frac{m_0}{q}\Big)$. This is due to the integration constant found by setting $t=0$ and then solving $z(0)=0$.

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Hint: $\int ln \, (1-at)dt=-\frac 1 a\int ln \, s \, ds$ where $s=1-at$ and $\int ln \, s \, ds=sln\, s -s+C$.

See my comment below for determining $C$.

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    $\begingroup$ That's what I did to get to the second term of my formula. Did I made it wrong? Mathematica gives the same results. $\endgroup$
    – Albert
    Aug 17, 2019 at 5:11
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    $\begingroup$ Your answer defers from the given answer only by a constant. Remember that when you integrate you have to add an arbitrary constant. The exact value of the constant can be determined by putting $t=0$: the initial velocity is $v_0$. $\endgroup$ Aug 17, 2019 at 5:22

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