Eksperimentti: hyppykorkeuden määrittäminen impulssilla: Difference between revisions
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The takeoff velocity is <math>v_0 = \frac Jm = \frac{ | The takeoff velocity is <math>v_0 = \frac Jm = \frac{209.25 Ns}{89.7 kg} = 2.33 m/s</math>. | ||
== Example 2: Zero the force plate == | == Example 2: Zero the force plate == | ||
Latest revision as of 18:16, 3 May 2022
Introduction

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
Note that , and thus the equation gives the correct equation.
Example
The example gives
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}m&=880N/9.81=89.7kg\\J&=464Ns-89.7kg\times 9.81\times 0.2895s=464Ns-254.75Ns=209.25Ns\end{aligned}}}
and thus we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}h&={\frac {J^{2}}{2gm^{2}}}\\&={\frac {(209.25Ns)^{2}}{2\times 9.81m/s^{2}\times (89.7kg)^{2}}}\\&={\frac {43785.5625}{315728.5716}}\\&=0.139m\\\end{aligned}}}
The takeoff velocity is Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle v_{0}={\frac {J}{m}}={\frac {209.25Ns}{89.7kg}}=2.33m/s} .
Example 2: Zero the force plate
References
https://www.thehoopsgeek.com/the-physics-of-the-vertical-jump/
https://www.brunel.ac.uk/~spstnpl/LearningResources/VerticalJumpLab.pdf