# Does the time to reach the highest point $P$ equal the time to reach the ground from $P$ in a trajectory, taking drag into account?

Assume a very simplified model without Corolis effect, the falloff of the local gravitational field and the like. My answer is no. It is sufficient to look at the vertical velocity of the projectile, because only that determines the time. The periods do not equal, because it is possible to vertically accelerate the projectile to a velocity which is greater than the terminal velocity $v_{term.}$ of the projectile in the medium. Therfore it may take longer for the projectile to come back down because it can only approach $v_{term.}$. I am not sure, though, how it behaves if the initial velocity is lower than $v_{term.}$. My guts say it behaves the same but a different explanation comes into effect. The projectile cannot reach the height which it would if there was no drag. Hence, the projectile cannot reach the initial velocity on the way back to the ground either and it takes longer.

The quadratic case is much more involved, but for a simple linear term describing drag the height of a projectile shot vertically is given by

$m \ddot{y} + \frac{g}{v_{T}}\dot{y}+g=0$

For graviational acceleration $g$, teminal velocity $v_T$ and mass $m$. Subject to the boundary conditions $y(t=0)$ and $\dot{y}(t=0)=v_0$ this differential equation has the solution

$y = \frac{m v_T}{g}(v_0 + v_T)(1-\exp({-\frac{g t}{m v_T}}))-v_T t$

The particle reaches the apex at $\dot{y}=0$, which after solving for $t$ is given by

$t_{max}= \frac{m v_T}{g} \log(1+\frac{v_0}{v_T})$

Thus we find at time $t=2 t_{Max}$ that

$y(t=2 t_{max}) = \frac{m v}{g}[v_0(v_0+2 v_T)-2 v_T(v_0+v_T) \log(1+\frac{v_0}{v_T})]$

if we find that at this time that $y$ is negative, we conclude the particle fell to ground from the apex quicker than it climbed to it. in term of the ratio $r=\frac{v_0}{v_T}$ this condition is

$\log(1+r) > \frac{r(r+2)}{2(r+1)}$

This is never satisfied and so we conclude that your intuition was correct. In fact these two quantities reach equality only for $r=0$ which is the pathological case of no motion.

I hope this walkthrough is able to offer you insight on how to tackle the more general problem.

• Thanks for that walkthrough. Just to be clear: It's enough to only take the 1D linear/vertical case into account, because it is possible to decompose the movement, right? – Lenar Hoyt Sep 3 '13 at 14:19
• For drag of the form described in the equation above, a vectorial equation does indeed decouple. If you take drag to be quadratic or any more complicated form, in general they do not. The drag form used above is Stoke's law, and is roughly speaking only valid for non-turbulent flow. – ComptonScattering Sep 3 '13 at 15:12
• I have only covered the drag equation for spheres consisting of a linear "viscous term" and a quadratic "pressure term" yet. Is this model not decomposable in two directions? What would be an intuitive explanation for that? – Lenar Hoyt Sep 5 '13 at 18:14
• In general a differential operator on a vector will not decouple into equations for each element. Such operators that do are known as linear operators. Occasionally it is possible to use a clever change of coordinates to decouple equations. Your viscous term is linear and so permits decoupling, whereas the pressure term is not. As for 'why not', it is hard for me to answer without knowing why it is that you would expect that they would. As for an attempt: It is because the force on the particle in the x-direction depends on its velocity in y. But I may be restating the obvious. – ComptonScattering Sep 6 '13 at 23:58