If the height of an upwardly thrown ball can be approximated by $h(t)=h_0+v_0t-\frac{g}{2}t^2$, what is $\frac{h(t_2)-h(t_1)}{t_2-t_1}$ conceptually?

Question

I know that it describes the secant through $(t_1,h(t_1))$ and $(t_2,h(t_2))$, but I wouldn't know how to interpret it in "real life" terms.

I also know that if $t_1$ tends towards $t_2$, I get the tangent at $t_2$, which would be the speed, so maybe the secant describes the average speed between the two points, but then there would exist average speeds of $0$, which doesn't make much sense to me.

(This was part of a recent math exam, which is what I took as justification to post it here.)

Answer 1

Your fraction is the slope of the secant through the involved points.

You could use it to apply the mean value theorem, which gives that there is some point in between, where secant slope and $\dot{h} = v_z$ have the same value.

And yes, the secant slope is the time average of $v_z(t)$ over $[t_1, t_2]$: $$ \bar{v}_z = \frac{1}{t_2-t_1} \int\limits_{t_1}^{t_2} v_z(t) \, dt = \frac{1}{t_2 - t_1} (h(t_2) - h(t_1)) $$

Answer 2

$h(t) = h_0 + v_0 - \frac{g}{2}t^2 $ is the equation that gives you the height that an object reaches, when you assume that the air has zero resistance on the object. This makes the situation really easy, since you are looking at a simple system of forces. Also, the existence of average speed of zero ($0$), means that the object that you are observing, had positive speed at $t_1$ and the exact same but negative speed at $t_2$, which gives you that the average speed on this interval was zero ($0$).