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 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 J = \int F dt = \Delta p = m\Delta v} . Actually 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 \Delta v} is our takeoff speed 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 v_1=0} , and 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 \Delta v = v_{0} = \frac{J}{m} = \frac1m \int F dt} . 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 s = s_0 + v_0 t + \tfrac12 at^2} and thus 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 h = v_0 t - \tfrac12 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 $v=v_{0}-gt$ and at the maximum height we have that $v=0$, and thus $v_{0}=gt$ and $t={\frac {v_{0}}{g}}$. 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 ${\begin{aligned}m&=880N/9.81=89.7kg\\J&=900Ns\end{aligned}}$

and thus 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 \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