## Re: Force required for a vertical jump

Jay,
An estimation of peak vertical jump force with assumptions or guesses marked with a (?) there are quite a few

We will call the displacement of the subject’s center of mass (CoM) in standing as zero. Let’s say that the displacement of the CoM at the point of take-off is raised 5cm (?) relative to standing. They will bend their legs when landing on the box, so let’s say this reduces required maximum CoM displacement by 25cm (?). The net vertical displacement of the CoM in flight is then:
66.9 – 5 – 25 = 36.9 cm

The vertical take-off velocity is: v = sqrt(2xgx0.369) = 2.69 m/s, and
Momentum at take-off is: P = 75x2.69 = 201.8 kg.m/s

The concentric (upwards) movement of a squat jump has gone from zero velocity to the take-off velocity. So the net impulse, the change in momentum, in this case is the final momentum = 201.8 N.s.
If this impulse was achieved in 0.5 sec (?) then the average force applied over this time period is: 403.6 N (impulse = force x time). This is not the peak force. Peak force may be twice (?) the average force in a squat jump (?). This does include the force due to gravity so we have to add on body weight (BW) to get total vertical force.
The peak force may therefore be around: [403.6 x 2] + BW = 807.2 + (75 x 9.81) = 1,543 N

For a counter movement jump (CMJ) the initial standing displacement of the subject CoM is still considered zero. There is an initial negative impulse relative to body weight followed by a positive impulse relative to body weight; the net impulse will still be equal to 201.8 N.s to achieve the same vertical CoM displacement of 36.9 cm. However, at the minimum body position force will be above body weight (it was equal to BW at the beginning of the squat jump), and with a constant impulse required (201.8 N.s) the difference between average force and peak force during the concentric phase will likely be less if the force is applied over the same time period (0.5 sec). It is likely that in the CMJ the time period is shortened.

If this same impulse was achieved in 0.3 sec (?) then the average force applied over this time period is: 672.7 N
The peak force may therefore be around: [672.7 x 1.5] + BW = 1009.0 + (75 x 9.81) = 1,745 N

It is all a bit of guess work to get average and peak force during the concentric phase. Knowing the duration of the concentric phase would help the estimation of average force; peak force however is still a fairly crude estimate and I am just guessing what it might be relative to average force during the concentric phase. I have not attempt to estimate the peak force during the eccentric phase of the CMJ.

Allan