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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}} $]]
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