Approximate impact force upon landing with parachute

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SUMMARY

The discussion focuses on calculating the approximate impact force experienced by a parachutist upon landing. The scenario involves a static line jump from 1000 feet, with a total weight of 300 lbs (136.078 kg) and a constant descent velocity of 24 ft/sec (7.3152 m/s). The participant utilized the impulse-momentum principle, calculating the momentum as 995.379 kg·m/s and estimating the impact force to be 1.99 kN (approximately 450 lbs) based on an assumed impact time of 0.5 seconds. This approach highlights the complexities of accurately determining landing forces in parachuting scenarios.

PREREQUISITES
  • Understanding of basic physics concepts such as momentum and impulse
  • Familiarity with the equations of motion, particularly F=m*a and p=m*v
  • Knowledge of static line parachuting and its mechanics
  • Ability to perform unit conversions between imperial and metric systems
NEXT STEPS
  • Research the physics of parachute dynamics and terminal velocity
  • Learn about the effects of different landing techniques on impact forces
  • Explore advanced calculations involving variable descent rates and parachute designs
  • Investigate safety measures and equipment used to mitigate landing forces in parachuting
USEFUL FOR

This discussion is beneficial for parachutists, military personnel involved in airborne operations, physics students, and anyone interested in the mechanics of landing forces in parachuting scenarios.

BCHurricane89
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Homework Statement


So this isn't a homework question, just something my brother-in-law asked me. He is in the Army, and wanted to know an approximate landing force he would experience when he lands. After attempting the problem, and looking things over, I realized it may be harder than once thought.

So, he told me he jumps from 1000ft (304.8m), and its a static line jump, so the parachute opens the instant he jumps from the plane. The total weight of him plus gear will be 300lbs (136.078kg), and he will fall at a CONSTANT velocity of 24ft/sec (7.3152m/s). From here, it was easy to calculate the time it would take him to reach the ground (41.667sec).


Homework Equations


Initially I tried calculating acceleration with dv/dt, then realized if he has a constant velocity, there would be no acceleration, and therefore my plan of simply using a F=m*a would not work.



The Attempt at a Solution


From here, I then decided maybe it's more of a impulse-momentum type question as I would need to approximate an impact time, which he said maybe 0.5 seconds. Of course to make this easier, I will assuming he lands completely in the vertical, and not at an angle. Obviously his legs and body are going to be absorbing the force, and to calculate this doesn't appear to be straightforward.

I turned to the momentum equation p=m*v, and calculated his momentum being 995.379kgm/s. For the Impulse equation of F*t=m*dv or rearranging for F=(m*dv)/t. Using the approximated imapce time of 0.5 seconds, this give a force of 1.99kN. I have no idea what to reference this too, or if it is even a correct way of approaching the problem, so any help would be appreciated. Thanks!
 
Physics news on Phys.org
1.99 kN = 1990 N = approx. 450 lbs. compared to a jump weight of 300 lbs.
 

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