NUMERICAL SOLUTIONS (Upvote will be given. Please answer the problem USING MS EXCEL. No long explanation needed. Take a screenshot of the formulas and results.) A. Determine the terminal velocity (maximum attainable free fall velocity) by iteration of equation (2)above.1. For initial time = 0, velocity = 0;2. Using v=0 as v(t), determine v(t+1) [For time = 1s, the velocity at ti+1 is 9.8m/s;3. Using v = 9.8 as the next v(t₁), determine v(t+1). Round off numbers to the nearest ten thousandth.4. Repeat the cycle so that the value v(ti+1) - v(ti) = 0.00005. Complete the values in the table below in a separate MS Excel file with file nameTime123456789v(t₁+1)9.800017.8012velocity (m/s)v(t₁)0.00009.8000time (s)v (t+1)-v(ti)9.80008.0012B. Graph the function with x-axis as the time (s) and the y-axis as the velocity (m/s) in the same MS Excelfile as problem A above.0.0000 Numerical Solution to the Falling Parachutist Problem:Fw=mgWhere:Object released from restmсVtiti+1FDFw=mgDuring accelerationA numerical iterative solution was formulated to be:FD=Fw=mgFw=mgA parachutist of mass 68.1 kg jumps out of a stationary hot air balloon. The drag coefficient is 12.5kg/s.Given the differential equation:(1)= acceleration due to gravity, 9.8 m/s²= mass of the parachutist, 68.1 kg=drag coefficient, 12.5 kg/s= velocity, in m/s= initial time interval, in seconds= following time interval, in secondsnet force=mg-mg = 0At terminal velocitydvdtFo-FwmV2v(ti+1)= v(ti) + [g-v(ti)](ti+11 - t₁)(2)

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Numerical Solution to the Falling Parachutist Problem: Where: DO E UD Fw=mg с Object released from rest A numerical iterative solution was formulated to be: V FD ti ti+1 Fw=mg During acceleration A parachutist of mass 68.1 kg jumps out of a stationary hot air balloon. The drag coefficient is 12.5kg/s. Given the differential equation: 9-12 v (1) g = acceleration due to gravity, 9.8 m/s² m = mass of the parachutist, 68.1 kg = drag coefficient, 12.5 kg/s = velocity, in m/s = initial time interval, in seconds = following time interval, in seconds FD=Fw=mg Fw=mg net force=Fp-Fw = mg-mg = 0 At terminal velocity v(ti+1)= v(t₁) + [g- / v(ti)](ti+1-t₁) (2)

Numerical Solution to the Falling Parachutist Problem: Where: DO E UD Fw=mg с Object released from rest A numerical iterative solution was formulated to be: V FD ti ti+1 Fw=mg During acceleration A parachutist of mass 68.1 kg jumps out of a stationary hot air balloon. The drag coefficient is 12.5kg/s. Given the differential equation: 9-12 v (1) g = acceleration due to gravity, 9.8 m/s² m = mass of the parachutist, 68.1 kg = drag coefficient, 12.5 kg/s = velocity, in m/s = initial time interval, in seconds = following time interval, in seconds FD=Fw=mg Fw=mg net force=Fp-Fw = mg-mg = 0 At terminal velocity v(ti+1)= v(t₁) + [g- / v(ti)](ti+1-t₁) (2)

Engineering Fundamentals: An Introduction to Engineering (MindTap Course List)

5th Edition

ISBN:9781305084766

Author:Saeed Moaveni

Publisher:Saeed Moaveni

Chapter8: Time And Time-related Variables In Engineering

Section: Chapter Questions

Problem 39P

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