View source for Eksperimentti: hyppykorkeuden määrittäminen impulssilla

== Introduction ==

[[File:ForcePlate jumping1.svg|thumb|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 ==

[[File:ForcePlate jumping2 Impulse Areas.svg|thumb]]

Impulse <math>J = \int F dt = \Delta p = m\Delta v</math>. Actually <math>\Delta v</math> is our takeoff speed because <math>v_1=0</math>, and we have <math>\Delta v = v_{0} = \frac{J}{m} = \frac1m \int F dt</math>.  Because <math>s = s_0 + v_0 t + \tfrac12 at^2</math> and thus we have <math>h = v_0 t - \tfrac12 gt^2</math> because <math>h_0=0</math> and <math>a=-g = -9.81m/s^2</math>. However, for the velocity we have <math>v = v_0 - gt</math> and at the maximum height we have that <math>v=0</math>, and thus <math>v_0 = gt</math> and <math>t=\frac{v_0}{g}</math>. Combining these two we have

<math>
\begin{align}
h &= v_0 t - \tfrac12 gt^2 \\
  &= v_0 \tfrac{v_0}g - \tfrac12 g\left( \frac{v_0}g \right)^2 \\
  &= \frac{v_0^2}{g} - \tfrac12 \frac{v_0^2}{g} \\
  &= \frac{v_0^2}{2g} \\
  &= \frac{J^2}{2gm^2} = \frac{1}{2g} \left( \frac Jm \right)^2
\end{align}
</math>

Note that <math>J/m = v_0</math>, and thus the equation gives the correct equation.

=== Example ===

The example gives

<math>
\begin{align}
m &= 880 N /9.81 = 89.7 kg \\
J &= 700 Ns - 89.7 kg \times 9.81 \times 0.2895 s = 700 Ns - 254.75 Ns = 445.25 Ns
\end{align}
</math>

and thus we have

<math>
\begin{align}
h &= \frac{J^2}{2gm^2} \\
  &= \frac{(445.25 Ns)^2}{2 \times 9.81 m/s^2 \times (89.7 kg)^2 } \\
  &= \frac{198 247. 5625}{315 728.5716} \\
  &= 0.63 m\\
\end{align}
</math>

The takeoff velocity is <math>v_0 = \frac Jm = \frac{445.25 Ns}{89.7 kg} = 4.96 m/s</math>.

== 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