# How do I justify why using the average velocity equation isn't useful when calculating projectile airtime where final height = initial height?

I've been trying to figure out different ways to explain to my students how to figure out the time in the air for a projectile and ran into the following issue:

When a projectile is thrown with an initial vertical velocity component and comes back down to the same height, the equation for average velocity (displacement / time = (vf+v0)/2) cannot be used. The equation is useful when the vertical displacement is nonzero, however.

Thus, I've been trying to figure out a clear, conceptual way to explain why this equation won't yield an answer for this specific scenario and am struggling. Can anyone help me out? Thank you!

Note that while I understand you can figure out the time to the max height and then double that time to get the answer, that avoids the issue that I'm trying to be able to explain.

• $\Delta x/\Delta t=(v_\text f+v_0)/2$ is always valid for 1D constant acceleration motion, so I am not sure why you are saying it is not valid here. Unless you only mean to say it is not useful? Commented Jul 17, 2021 at 1:31
• If my goal is to solve for the time the projectile is in the air, I don't see how that equation helps me solve for t in the case that the object's final height is equal to its initial height. If I should be arguing that it's "not useful" instead, then how do I explain that conceptually as opposed to mathematically? Perhaps another way of asking my question is why is this equation limited in usefulness? (Also, thank you for the response! My first time posting on this forum) Commented Jul 17, 2021 at 16:21

" the equation for average velocity (displacement / time = (vf+v0)/2) cannot be used." (to determine the time of flight when $$v_f = -v_o$$)

is true and it's because there are many possible times, if the acceleration isn't known.

The answer for the time depends on the acceleration - imagine a projectile sent from the ground on the moon, compared to from the ground on earth.

The $$v_f$$ and $$v_o$$ could be the same, the average displacement would be the same, but the projectile would have travelled higher and taken longer, on the moon, before returning to ground level. There is no unique answer for the time if the acceleration isn't known. That's why the equations can't give a proper result.

• could you explain please what quantities are known, are they $v_0$, $v_f$ and $a$? If $a$ is known use $t=\frac{\Delta v}{a}$. If $a$ isn't known it's not possible to find the time when $v_f = - v_0$ and that's why it won't give an answer in that situation Commented Jul 17, 2021 at 18:31