Question

7. Calculate the height of the center of mass above its starting height during a squat jump based on the following information: BW= 670 N, total vertical ground reaction force= 788 N, time of force application= 0.9s.

Answer 1

Answer:

h = 12.6 cm

Explanation:

given,

ground reaction = 788 N

time of force application = 0.9 s

BW = 670 N

Net force = 788 - 670

               = 118 N

impulse = F x t

I = 118 x 0.9

I = 107.1 N s

impulse is equal to change in momentum

$v = \dfrac{I}{m}$

$v =\dfrac{107.1}{\dfrac{670}{9.8}}$

       v = 1.57 m/s

v is the initial velocity before jump

now, height of center of mass

using equation of motion

v² = u² - 2 g h

$h = \dfrac{u^2}{2g}$

$h = \dfrac{1.57^2}{2\times 9.8}$

     h = 0.1257 m

     h = 12.6 cm

the height of center of mass is equal to h = 0.126 m or 12.6 cm