In this experiment, you will deduce your jumping speed using standard gravity.

Standard gravity [[$ g $]]​, change in velocity [[$ \Delta v $]]​ ja and time period [[$ \Delta t $]]​ are related by the equality
​[[$$ g = \frac{\Delta v}{\Delta t}. $$]]​
From this, we may solve
​[[$$ \Delta v = g \Delta t. $$]]​
If you jump upwards at speed [[$ v_{\text{hyppy}} $]]​, you will hit the ground at a velocity that is equal in value but opposite in direction. Mathematically, your speed at impact is [[$ -v_{\text{hyppy}} $]]​. Therefore, the change in your velocity is
​[[$$ \Delta v = 2 v_{\text{hyppy}} .$$]]​
By combining these pieces of information, we obtain
​[[$$ 2 v_{\text{hyppy}} = \Delta v = g \Delta t, $$]]
and this leads to the equality
​[[$$ v_{\text{hyppy}} = \frac{g \Delta t}{2}. $$]]

To find your jump speed, you only need to measure the time you spend in air during your jump. It follows that the experiment can be carried out with a stopwatch only.

My jump took [[$ \Delta t = $]] s, so my jump velocity is
[[$ v_{\text{hyppy}} = \frac{g\Delta t}{2} = \frac{10 \text{ m}\text{s}^{-2}}{2}\ \cdot $]] s = [[$\frac{\text{m}}{\text{s}} $]]