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picture from Wikipedia page about projectile motion, under air resistance section

Wikipedia: Projectile motion - Derivation of the time of flight:

$$c_1t+c_2+c_3e^{c_4t}=0$$

I understand how the equation for $y(t)$ is of the form that they describe. From my understanding, $W(te^t) = t$, so we want the expression in front of the $e$ to be the same as the power in order to solve for $t$. However, I am not sure how they actually transformed the expression for $y(t)$ into one that can be solved by the Lambert W function. What algebraic steps were taken?

IV_
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Samwit 7
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  • photos and links are discouraged because they die. Math can be rendered with mathjax. The community typically requires an explanation of what you've tried and specifically what your problem is to provide any help – kevinkayaks Aug 24 '23 at 13:46
  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Aug 24 '23 at 13:52

1 Answers1

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$$c_1t+c_2+c_3e^{c_4t}=0$$ $$c_3e^{c_4t}=-c_1t-c_2$$ $$c_3=(-c_1t-c_2)e^{-c_4t}$$

We see, both the factor and the exponent of the exponential term are linear functions of the solution variable ($t$). Such equations can be solved in terms of Lambert W.

We have to try to get the same expression for the factor and the exponent.

$$(-c_1t-c_2)e^{-c_4t}=c_3$$ $$(-c_1t-c_2)e^{-c_4t-\frac{c_2c_4}{c_1}}=c_3e^{-\frac{c_2c_4}{c_1}}$$

Expanding the exponent:

$$(-c_1t-c_2)e^{\frac{(-c_1t-c_2)c_4}{c_1}}=c_3e^{-\frac{c_2c_4}{c_1}}$$ $$\frac{(-c_1t-c_2)c_4}{c_1}e^{\frac{(-c_1t-c_2)c_4}{c_1}}=\frac{c_3c_4}{c_1}e^{-\frac{c_2c_4}{c_1}}$$

Now we have the same expression for the factor and the exponent, and we can use Lambert W therefore ($xe^x=c\implies x=W(c)$):

$$\frac{(-c_1t-c_2)c_4}{c_1}=W\left(\frac{c_3c_4}{c_1}e^{-\frac{c_2c_4}{c_1}}\right)$$ $$t=-\frac{1}{c_4}W\left(\frac{c_3c_4}{c_1}e^{-\frac{c_2c_4}{c_1}}\right)-\frac{c_2}{c_1}$$

IV_
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