Velocity of projectile with both quadratic and constant resistive force Suppose we have a projectil

Velocity of projectile with both quadratic and constant resistive force
Suppose we have a projectile of mass m that impacts a material with velocity ${v}_{0}$. As it travels through the material it is subject to the resistive force $F=\alpha {v}^{2}+\beta$. How can I determine $v\left(t\right)$, the velocity of the projectile at time t where t=0 is the impact time, and $v\left(p\right)$, the velocity of the projectile when it reaches a penetration depth of p?
This question addresses v(t) for an exclusively quadratic resistive force ($\beta =0$), but is there a more general formula for when there is a constant component?
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Gallichi5mtwt
Using $vdv=adx$ where $a=F/m=-\frac{\alpha }{m}{v}^{2}-\frac{\beta }{m}$ we have
$vdv=\left(-\frac{\alpha }{m}{v}^{2}-\frac{\beta }{m}\right)dx\to \int \frac{v}{\frac{\alpha }{m}{v}^{2}+\frac{\beta }{m}}dv=-\int dx$
Defining the new variable $u=\frac{\alpha }{m}{v}^{2}+\frac{\beta }{m}$ we have $\frac{du}{dv}=2\frac{\alpha }{m}v$. Thus, $vdv=\frac{m}{2\alpha }du$ and the integral reduces to
$\frac{m}{2\alpha }\int \frac{1}{u}du=-\int dx$
Subsequently,
$\frac{m}{2\alpha }\mathrm{ln}u=-x+C$
We use the initial condition that for x=0, $v={v}_{0}$ and $u={u}_{0}=\frac{\alpha }{m}{v}_{0}^{2}+\frac{\beta }{m}$. Hence $C=\frac{m}{2\alpha }\mathrm{ln}{u}_{0}$. With some straightforward algebra you should be able to get
$u\left(x\right)={u}_{0}\mathrm{exp}\left(-\frac{2\alpha }{m}x\right)$
Let x=p and you'll get u(p). From that you get v(p).