Simple sum of forces parachute problem

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SUMMARY

The discussion centers on calculating the average force exerted on a 64 kg parachutist who falls at a speed of 6.3 m/s and comes to a stop after impacting the ground over a distance of 0.92 m. The user correctly applies the kinematic equation (Vf)^2 = (Vi)^2 + 2a(delta Y) to find the acceleration, resulting in a value of -21.57 m/s². Subsequently, using F = ma, the calculated force is -1402 N, indicating the force is indeed directed upward, counteracting the downward motion of the parachutist.

PREREQUISITES
  • Understanding of kinematic equations in physics
  • Knowledge of Newton's second law (F = ma)
  • Familiarity with concepts of force and acceleration
  • Basic understanding of vector directions in physics
NEXT STEPS
  • Explore advanced kinematic equations for varying acceleration scenarios
  • Study the implications of force direction in collision physics
  • Learn about energy conservation during free fall and impact
  • Investigate real-world applications of force calculations in parachuting
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of free fall and impact forces, particularly in the context of parachuting and similar scenarios.

kirby27
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a 64 kg parachutist falling vertically at a speed of 6.3 m/s impacts the ground, which brings him to a complete stop in a distance of .92 m. assuming constant acceleration after his feet first touch, what is the average force exerted on the parachutist by the ground?

i know:
Vf=0
Vi=6.3
delta Y = .92

i used (Vfy)^2=(Viy)^2+2a(deltaY)

and found a = -21.57

i then used F=ma: F= (65)(-21.57)

and got -1402 N

is this right?
 
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Correct method.

Is the force exerted by the ground upward, or is it downward?
 
i think the force exerted by the ground on the man is upward? should my answer be positive?
 
I would think so.
 

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