Lab4

$$F = m\cdot a,$$ so we can substitute $$a = \frac{\Delta v}{\Delta t}$$ to get $$F = \frac{m\cdot\Delta v}{\Delta t},$$ and rearrange to $$F \Delta t = m \Delta v,$$ so impulse equals momentum. We can check that the units match up with $$\begin{align*} \text N \cdot \text s &= \text{Kg} \cdot \text m \cdot \text s^{-1} \\ \end{align*}.$$ Now we can calculate take-off velocity with $$\Delta v = \frac{F \Delta t}{m}.$$ We can then use velocity to calculate kinetic energy with $$KE = \frac12mv^2.$$ Since this will be converted to gravitational potential energy during the jump, we will have $$KE = \frac12mv^2 = GPE = mg \Delta h.$$ substituting $g = 9.81$ and dividing my the mass gives $$v^2/2 = 9.81 \Delta h$$ which rearranges to $$\Delta h = v^2/19.62.$$