I've been told that I can calculate how high some one is jumping on a trampoline by timing 10 bounces I am attending a school (in Oakland CA), taking trampoline classes, none of the trainers are aware of the equation used for determining height jumped. I would guess this is a simple question for this exchange but I am hoping someone is willing to help out.
 A: In very simplistic terms the total time per jump cycle is broken down to two parts.


*

*Time spent in the air decelerating downwards by $g$ 

*Time spent in contact with the trampoline accelerating upwards by $a$


The total time $T$ is
$$ T = \frac{2 v}{a} + \frac{2 v}{g} $$
where $v$ is the launch velocity (equalling the impact velocity). The total jump height can be estimated from the launch velocity $$ h = \frac{v^2}{2 g}$$
Combined the above give us an estimate of the height based on time only if the acceleration $a$ is known, as well as the total time $T$ and gravity $g$.
$$ h = \frac{T^2 a^2 g}{8 (a+g)^2} $$
In this post the acceleration is measured from video data for a particular kind of trampoline.
Using this data of $a=26.6\;\mathrm{\frac{m}{s^2}}$, $g=9.81\;\mathrm{ \frac{m}{s^2}}$ you get
$$ h\mbox{ [meters]} = 0.621 (T \mbox{ [seconds]})^2 $$

As an example, if the total time is $T=1.95\,{\rm s}$ then the height should be $$ h = 0.621\cdot (1.95)^2 = 2.36\,{\rm m} $$
A: While you can do this by timing n bounces, it's tricky to separate the time of the rebound separate from the air time.
I ran into an article online that considered the rebound part to be a simple harmonic oscillator.  Some tests showed that while the time to do 60 bounces at just barely perceptable movement up to the point where feet were about to leave the mat it was indeed, a very uniform time.  (within a second) when I used this time as bounce time later I got absurd results.
The return force on a trampoline is messy.

*

*As the displacement increases are resistance becomes a significant factor.

*When the mat is close to level, the tension on the mat nearly cancels out, as the mat is deformed, the vector cancellation decreases.  Force increases as the sin of the angle of the cone of depression.  So far, all in accord with SHM.  But also you are stretching the springs, so there is a multiplier.  I think for first order approximation, that the force goes up as the square of the angle of depression.

However, there is an easier way.
Video it with your phone on a slo-mo setting.  Now time some number of jumps, while videoing it. I suggest using at least 30 to minimize start/stop timing errors.
Analyze the vid counting frames for several sample bounces, and the subsequent air time.
Example:  YOur 60 jump time is 80 seconds.  Counting frames, the bounce takes 16 and the jump 24.  Use ratio and proportion, this means you spent 48 secons in the air and 32 on the mat.
48 seconds is 0.8 seconds per jump.  Thats Up and Down time, so use half that time 0.4 seconds for the falling time.  $$d = {1\over2} at^2$$
$$d= {1\over2} * 10 * 0.4^2$$
$$d = 0.8 m$$
