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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(t), 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) = 0.0000 5. Complete the values in the table below in a separate MS Excel file Time 1 2 3 4 5 6 7 8 9 ... v(t₁+1) 9.8000 17.8012 velocity (m/s) v(t₁) 0.0000 9.8000 time (s) v (t+1)-v(t₁) 9.8000 8.0012 B. Graph the function with x-axis as the time (s) and the y-axis as the velocity (m/s) in the same MS Excel file as problem A above. 0.0000

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(t), 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) = 0.0000 5. Complete the values in the table below in a separate MS Excel file Time 1 2 3 4 5 6 7 8 9 ... v(t₁+1) 9.8000 17.8012 velocity (m/s) v(t₁) 0.0000 9.8000 time (s) v (t+1)-v(t₁) 9.8000 8.0012 B. Graph the function with x-axis as the time (s) and the y-axis as the velocity (m/s) in the same MS Excel file as problem A above. 0.0000

College Algebra
College Algebra
10th Edition
ISBN:9781337282291
Author:Ron Larson
Publisher:Ron Larson
Chapter3: Polynomial Functions
Section3.5: Mathematical Modeling And Variation
Problem 7ECP: The kinetic energy E of an object varies jointly with the object’s mass m and the square of the...
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UPVOTE WILL BE GIVEN. COMPLETE THE TABLE A USING MS EXCEL.
Numerical Solution to the Falling Parachutist Problem: Fw=mg Where: Object released from rest m с V ti ti+1 FD Fw=mg During acceleration A numerical iterative solution was formulated to be: FD=Fw=mg Fw=mg 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: (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 seconds net force =mg-mg = 0 At terminal velocity dv dt Fo-Fw m V2 v(ti+1)= v(ti) + [g-v(ti)](ti+11 - t₁) (2)
Transcribed Image Text:Numerical Solution to the Falling Parachutist Problem: Fw=mg Where: Object released from rest m с V ti ti+1 FD Fw=mg During acceleration A numerical iterative solution was formulated to be: FD=Fw=mg Fw=mg 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: (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 seconds net force =mg-mg = 0 At terminal velocity dv dt Fo-Fw m V2 v(ti+1)= v(ti) + [g-v(ti)](ti+11 - t₁) (2)
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) = 0.0000 5. Complete the values in the table below in a separate MS Excel file Time 1 2 3 4 5 6 7 8 9 v(t₁+1) 9.8000 17.8012 velocity (m/s) v(t₁) 0.0000 9.8000 time (s) v (t+1)-v(ti) 9.8000 8.0012 B. Graph the function with x-axis as the time (s) and the y-axis as the velocity (m/s) in the same MS Excel file as problem A above. 0.0000
Transcribed Image Text: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) = 0.0000 5. Complete the values in the table below in a separate MS Excel file Time 1 2 3 4 5 6 7 8 9 v(t₁+1) 9.8000 17.8012 velocity (m/s) v(t₁) 0.0000 9.8000 time (s) v (t+1)-v(ti) 9.8000 8.0012 B. Graph the function with x-axis as the time (s) and the y-axis as the velocity (m/s) in the same MS Excel file as problem A above. 0.0000
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