Pad 8:19 PM ??@ 37%.- ? usi38ok.theexpertta.com I Help Homework 6 Begin Date: 3/

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Pad 8:19 PM ??@ 37%.- ? usi38ok.theexpertta.com I Help Homework 6 Begin Date: 3/9/2018 12:00:00 AM--Due Date: 3/28/2018 11:59:00 PM End Date: 3/28/2018 11:5900 PM (6%) Problem 17: Mike has a mass of m M 97 kg. He Jumps out of a perfectly good airplane that is 2000 m above the ground. After he falls 1000 m, when his downward speed is v 76 m/s, Mike opens his parachute. The positive y-direction is downward. Randomized Variables MIM 97 kg v-76 m/s nt Status ere for d view ? 33% Part (a) Calculate the average magnitude of the upward force, F , in Newtons of the air resistance on Mike during his initial descent. Grade Summary Status Completed Completed Completed Completed Completed Completed Posential sinO cotan0 coso asino atanacotan0 sinh0 tano acoso 4 56 (0% per attempt detailed view cosh0 tanhcotanh0 O Degrees Radians Hint Completed Completed Completed Hints: 2% deduction per hint. Hints remaining:- Feedback: 2 dedaction per feedback 33% Part (b) After Mike opens his parachute, he continues to descend, eventually reaching the ground with a speed of 4.0 m/s. Calculate the average upward force, Fa in Newtons, during this part of Mike's descent Completed Completed Partial Partial Partial 33% Part (c) At the same time Mike jumps out ofthe airplane, his wallet (mass ofm, 3 kg) falls out of his pocket. Calculate the wallet's downward speed, Vuyin m's, when it reaches the ground. For this calculation, assume that air resistance is negligible. Allonen. © 2018 Espen TA, LLC

Explanation / Answer

Here we have

velocity of mike after 1000 m descent = 76 m/s

let deceleration due to air resistance = aL

by third law of motion we have:

v^2 - u^2 = 2*(g-aL)*s

here u = 0

76*76 = 2*(9.8-aL)*1000

9.8 - aL = 2.89

aL = 6.91 m/s^2

Upward force, Fl = mass * aL = 97*6.91 = 670.27 N (two significant digits) ~ 670.3 N (one significant digit)

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