I am struggling to figure out how to calculate the final velocity of a projectile that is shot vertically upwards and then let fall down again to the starting point. I can easily calculate this in multiple ways when air resistance is not a factor and obviously in that case the end velocity will be the same as the starting one but in the opposite direction. What formulas do I need to use when air resistance becomes relevant?
I would appreciate a quick walk-through for say an object shot upwards at 20m per s with a weight of 1kg and constant gravity of 9.8. I don't care about getting the air resistance correct and like to simply use a coefficient like 0.1 for a given object.
Best Answer
Assume that a particle is shot upwards with an initial velocity of $v_0$ at $t = 0$. It then experiences a linear drag force of the form $F_\text{drag} = - \alpha v$, where $\alpha$ is a constant of proportionality. The equation of motion is then $$ F_\text{grav} + F_\text{drag} = ma \quad \Rightarrow \quad - m g - \alpha v = m \frac{dv}{dt} \quad \Rightarrow \quad \frac{dv}{dt} = -g - \frac{\alpha}{m} v. $$ This is a separable differential equation, and can be re-arranged and integrated: \begin{align*} - \frac{dv}{g + \alpha v/m} &= dt \\ - \frac{m}{\alpha} \int_{v_0}^v \frac{dv'}{mg/\alpha + v'} &= \int_{0}^t dt' \\ \ln \left[ v' + \frac{mg}{\alpha} \right]_{v_0}^v &= - \frac{\alpha t}{m} \\ \frac{v + mg/\alpha}{v_0 + mg/\alpha} &= e^{-\alpha t/m} \end{align*} $$ \boxed{v(t) = \left( v_0 + \frac{mg}{\alpha} \right) e^{-\alpha t/m} - \frac{mg}{\alpha} } $$ The solution for $x(t)$ can be then be found by integrating this once more with respect to time. This method can also be used to find the behavior of projectiles under more realistic forms of drag force, such as quadratic drag; the integrals are just a bit harder.
Alternately, if you a relationship between $v$ and $x$, you can use the chain rule to re-write the above equation as $$ \frac{dv}{dx} \frac{dx}{dt} = - g - \frac{\alpha}{m} v \quad \Rightarrow \quad \frac{dv}{dx} v = - g - \frac{\alpha}{m} v $$ Following the same logic, we then get \begin{align*} - \frac{m}{\alpha} \int_{v_0}^v \frac{v' dv'}{mg/\alpha + v'} &= \int_{0}^x dx' \\ \left[ v' - \frac{mg}{\alpha} \ln \left( v' + \frac{mg}{\alpha} \right) \right]_{v_0}^v &= - \frac{\alpha x}{m} \\ v - v_0 - \frac{mg}{\alpha} \ln \left( \frac{ v + mg/\alpha}{v_0 + mg/\alpha} \right) &= - \frac{\alpha x}{m} \end{align*} This equation can't be solved exactly to obtain $v$ as a function of $x$ (because of the presence of both logarithmic and polynomial terms), but you could figure out the speed at any height using numerical root-finding techniques; you could also solve for the maximum height by setting $v = 0$ and finding the corresponding $x$. And, as before, this technique could be adapted to other forms of the drag force such as quadratic drag.