I was still struggling with the physics of PK. I asked some of my friends:

> So, math/physics people: If you have a 75 kg object falling at 9.8

> m/s^2 [gravity] what is the time till impact when dropped from 1, 2,

> 3m? What is the force of impact? If you need a decelleration time, try

> .1 sec. The answers people are giving me are far from what I expected,

> and I'm struggling through online physics tutorials to try to figure

> it out for myself.

>

> Then the reverse: What force at what accelleration is needed to move

> the object up? Let's say .5, 1 and 1.5 m, since I am a white man, and

> cannot jump

Dave wrote:

s = 0.5 a t^2

t = sqrt( 2s / a )

v = at

P = mv

F = dP/dt

t1 = sqrt( 2/9.8 ) = 0.45s v1 = 4.41m/s P1 = 331N-s

t2 = sqrt( 4/9.8 ) = 0.64s v2 = 6.27m/s P2 = 470N-s

t3 = sqrt( 6/9.8 ) = 0.78s v3 = 7.64m/s P3 = 573N-s

Since all the momentum is given up in the inelastic collision, the force

of impact is, as you thought, just the momentum above divided by the

amount of time you assume for the deceleration.

Here's my thoughts on finding that time: when the object hits the

ground, the leading edge comes to a stop immediately, but the higher

parts are still falling. That means there's a compression wave going

through the object. The time it takes for the compression wave to go

from the bottom to the top must essentially be the deceleration

time--and this should just be the sound speed in the medium times the

height of the object.

As for moving things up, we have the same problem as above: for a short

interval, acceleration must be greater than g, but we don't know the

interval, and hence not the force. But if it was you jumping, think

about the time it takes for you to go from a knees flexed position to a

knees straight, at which point you leave the ground. Only during that

interval can you be exerting any force at all, and you might as well

assume it's constant. After that, it's just ballistics.

Dave