Eksperimentti: hyppykorkeuden määrittäminen impulssilla

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 $J=\int Fdt=\Delta p=m\Delta v$. Actually $\Delta v$ is our takeoff speed because $v_{1}=0$, and we have $\Delta v=v_{0}={\frac {J}{m}}={\frac {1}{m}}\int Fdt$. Because $s=s_{0}+v_{0}t+{\tfrac {1}{2}}at^{2}$ and thus we have $h=v_{0}t-{\tfrac {1}{2}}gt^{2}$ because 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 h_0=0} and 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 a=-g = -9.81m/s^2} . However, for the velocity we have 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 v = v_0 - gt} and at the maximum height we have that 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 v=0} , and thus 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 v_0 = gt} and 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 t=\frac{v_0}{g}} . Combining these two we have

${\begin{aligned}h&=v_{0}t-{\tfrac {1}{2}}gt^{2}\\&=v_{0}{\tfrac {v_{0}}{g}}-{\tfrac {1}{2}}g\left({\frac {v_{0}}{g}}\right)^{2}\\&={\frac {v_{0}^{2}}{g}}-{\tfrac {1}{2}}{\frac {v_{0}^{2}}{g}}\\&={\frac {v_{0}^{2}}{2g}}\\&={\frac {J^{2}}{2gm^{2}}}\end{aligned}}$

Example

The example gives Failed to parse (syntax error): {\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