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$\bullet$$\bullet$ A 65.0 -kg parachutist falling vertically at a speed of6.30 $\mathrm{m} / \mathrm{s}$ impacts the ground, which brings him to a completestop in a distance of 0.92 $\mathrm{m}$ (roughly half of his height).Assuming constant acceleration after his feet first touch theground, what is the average force exerted on the parachutist bythe ground?

1402 $\mathrm{N}$

Physics 101 Mechanics

Chapter 5

Applications of Newton's Law

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

Rutgers, The State University of New Jersey

University of Washington

Hope College

Lectures

04:01

2D kinematics is the study…

03:28

Newton's Laws of Moti…

02:04

When starting a foot rac…

03:15

A 65.0-kg basketball playe…

04:18

After falling from rest fr…

04:06

A parachutist relies on ai…

01:35

Integrated Concepts When s…

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Predict/Calculate $\mathrm…

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A skydiver, who weighs $65…

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A 60.0 -kg male dancer lea…

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our friend a situation. We choose the upward direction to be positive and forgiven that initial velocity is 6.2 meters per second. Squared downwards. That's negative. 613 uh, meters for second. Excuse me on then. And the height, Uh, that's no pressure has fallen is 0.90 meters and that's fall downwards. That's the negative again so we can find acceleration by off the equation of motion given as b squared equals C squared plus two. Why? Turns out why and so a Y acceleration, light elections just be squared. Might spin out squared where two times started Why, uh, the squared zero come rest. Finally. So you have negative negative 6.2 meters per second square over two times negative 90 meters on. So we get 21.6 years per second squared acceleration on, uh, on the upward direction s So now way. Use Newton's second law, which says that the net force in the white direction should be equal to mass times acceleration. Why direction? And so we know the right hand side. Massive 65 kilograms coloration. There's 21.6 meters per second squared and sell. This gives us a net force of 1,440. All right, so when the parachutist just touching the ground. But there's a force exerted by the ground on the parachuters. A cz well as the force of gravity now the force of gravity acting on the parachutists. So again, you second law. But this time for the Y direction. So we have met forces 14 or four. Well, still the wind direction. But now we have met, forcing way have met force and we want force exerted by the ground f ground. So this will be f ground minus the force of gravity of fact downwards. And so f ground will be 40 no for class mg mg a 65 times 9.8 14 for there. And so that gives us the force exerted by the ground off 2,041 years.

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