# Force and Impulse when Landing

1. Apr 4, 2013

### Killic

1. The problem statement, all variables and given/known data
When a person parachutes, the impact velocity is equal to that attained in free fall from a height of 4.5 m. After contacting the ground, the jumper’s momentum is quickly brought to zero by the earth. Assuming that the person has a mass of 80 kg, what is the average force on the jumper’s feet if he lands stiff-legged and the impulse lasts only 0.019 s?

3. The attempt at a solution
v22 = v12 + 2aΔd
v22 = (2)(1.8)(4.5)
v2 = 9.4 m/s [down]

p = mv
p = (90)(9.4)
p = 752 N

F = Δp/Δt
F = 750/0.019
F = 3.9 × 104 N [up]

However, the correct answer is 4.03 x 104 N [up]. My teacher hinted towards the normal force to the person's weight is added to the answer. This confuses me as I thought that the impulse takes that into consideration already. Someone please help clarify this for me. Thanks in advance.

2. Apr 5, 2013

### Staff: Mentor

Hi Killic... http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

The answer you obtained (through omitting gravity) is the answer that applies if the problem was reworded for horizontal motion, viz., a speed skater comes off the rink at 9.4 m/sec and collides with a plasterboard-encased concrete wall, coming to a stop in 0.019 secs.

But it takes additional force to halt a body moving vertically because gravity is trying to accelerate the body (pulling downwards) all the time that friction (pushing upwards) is trying to slow the body.

Last edited by a moderator: May 6, 2017
3. Apr 5, 2013