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?
$$c_1t+c_2+c_3e^{c_4t}=0$$ $$c_3e^{c_4t}=-c_1t-c_2$$ $$c_3=(-c_1t-c_2)e^{-c_4t}$$
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 ($xe^x=c\implies x=W(c)$):
$$\frac{(-c_1t-c_2)c_4}{c_1}=W\left(\frac{c_3c_4}{c_1}e^{-\frac{c_2c_4}{c_1}}\right)$$ $$t=-\frac{1}{c_4}W\left(\frac{c_3c_4}{c_1}e^{-\frac{c_2c_4}{c_1}}\right)-\frac{c_2}{c_1}$$