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(III) (a) Calculate the impulse experienced when a $65-\mathrm{kg}$ person lands on firm ground after jumping from a height of 3.0 $\mathrm{m}$ . (b) Estimate the average force exerted on the person's feet by the ground if the landing is stiff-legged, and again (c) with bent legs. With stiff legs, assume the body moves 1.0 $\mathrm{cm}$ during impact, and when the legs are bent, about 50 $\mathrm{cm} .$ [Hint. The average net force on her which is related to impulse, is the vector sum of gravity and the force exerted by the ground.]

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a) $5.0 \times 10^{2} k g . m / s,$ upwardsb) $1.9 \times 10^{5} m / s,$ upwardsc) $4.5 \times 10^{3} \mathrm{m} / \mathrm{s}$ , upwards

Physics 101 Mechanics

Chapter 9

Linear Momentum

Motion Along a Straight Line

Kinetic Energy

Potential Energy

Energy Conservation

Moment, Impulse, and Collisions

University of Michigan - Ann Arbor

Hope College

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it's part of this problem requires impulse calculation on dso for to find impulse, which is mass times change in velocity. You need velocity, right? Flossy Like given you get that from equating firm energy conservation. Ah, kinetic energy is equal to potential energy. Right? So, um, 1/2 has mass times final velocity squared is equal to mass times gravitational acceleration, times, height, and so velocity is just, um uh is just two times exceptional to gravity. 9.8 meters per second. Squared times height, three meters above ground level. And that is 7.67 meters per second. Um, and so right. And, um so this is part of a, uh, and so impulse will just be mass times the velocity and this will be S O. J is equal to m times Delta v So mass times a change of velocity, I should say So mess the 65 kilograms times 7.6 minus zero, which is just 7.67 meters per second. So this gives us an impulse off 4 98 uh, occasion meters per second, close to 500. And since this is positive, we're taking upward to re positive. It is, um, upwards and is an airport's acting impulse. Uh, in part B. We want to use Newton's second law, which is that the net force is equal to mass times acceleration. Let's move to the next page. So that force is force exerted by the ground, uh, on the person's feet. So after two minus force of gravity from the exerted by the person on the ground is equal to mass times acceleration again upwards, its positives of this is positive. And this is negative. Ah, f g is what we want. And so that's just mass times acceleration plus crab. So gravity, we know is 9.8 acceleration. On the other hand, um, is, uh, found from the equations of motion. So we use v squared. We are the use of the square, the square, the final squared, uh, equals V initial squared over, too. Time, Stelter. Ex change in possession equals acceleration. Right? So there's a 6.766 point 7.67 Excuse me, uh, meters per second beers per second. Quantity squared minus zero. Of course. Over. Two times for the stiff like we have 10 centimeters. Appoint one meters. Um, okay. And, um And so this gives us to 94.1 to 94.1 meters per second. Squared quick. Uh, therefore, uh, f g is just 65 kilograms times to 94.1 plus 9.8 meters per second squared, giving us force exerted by the ground of 1.9 times 10 to the five mutants. Okay. And finally in part. See, we're using the same formula F two calls em time safe was t a is different here. So a will be a 7.67 meters per second quietly squared over this time two times 50 centimeters or 500.5 meters. Okay, so that gives us 58 point in me meters per second squared. Awesome. Therefore, forced from the ground is just 65 times 50 repointing really plus 9.8. Minister per second squared, giving us 4.5 times 10 to the three mutants

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