Eksperimentti: hyppykorkeuden määrittäminen impulssilla: Difference between revisions

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<math>
<math>
\begin{align}
\begin{align*}
h &= \frac{J^2}{2gm^2} \\
h &= \frac{J^2}{2gm^2} \\
   &= \frac{(900 Ns)^2}{2 \times 9.81 m/s^2 \times (89.7 kg)^2 } \\
   &= \frac{(900 Ns)^2}{2 \times 9.81 m/s^2 \times (89.7 kg)^2 } \\
   &= \frac{810000}{8046.09} \\
   &= \frac{810000}{8046.09} \\
   &= 100.67 m\\
   &= 100.67 m\\
\end{align}
\end{align*}
</math>
</math>



Revision as of 17:46, 3 May 2022

Introduction

Force exerted on the force plate

Jumping on the force plate you can feel the force. We use time of flight method to estimate the height of the jump.

Theory

Impulse . Actually is our takeoff speed because , and we have . Because and thus we have because and . However, for the velocity we have and at the maximum height we have that , and thus and . Combining these two we have

Example

The example gives Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align*} m &= 880 N /9.81 = 89.7 kg \\ J &= 900 Ns \end{align*} and thus we have <math> \begin{align*} h &= \frac{J^2}{2gm^2} \\ &= \frac{(900 Ns)^2}{2 \times 9.81 m/s^2 \times (89.7 kg)^2 } \\ &= \frac{810000}{8046.09} \\ &= 100.67 m\\ \end{align*} }

References

https://www.thehoopsgeek.com/the-physics-of-the-vertical-jump/

https://www.brunel.ac.uk/~spstnpl/LearningResources/VerticalJumpLab.pdf