Task 2 code: height = 4000; %initial height u = 0; %initial velocity g = 9.81; %

Question

Task 2 code:

height = 4000; %initial height

u = 0; %initial velocity

g = 9.81; %gravitational velocity

t = 0;

ts = 0.1; %time step

n = 1;

while height > 0

  

if t < 60

h = u*ts; %height at that specific time

v = (g-(1/6)*u)*ts; %formula for velocity at that specific time

u = u+v; %new initial velocity

height = height-h; %keep looping until the height reaches 0

t = t+ts; %current time increase time step

else

v = (g-(5/3)*u)*ts; %velocity when parachute is opened

h = u*ts; %height at that specific time

u = u+v; %new initial velocity

height = height-h; %keep looping until the height reaches 0

t = t+ts; %current time increase time step

end

  

h1(n) = height; %the height at each time step

v1(n) = u; %the velocity at each time step

n = n+1;

end

%plotting

t1 = ts*(1:n-1);

plot(t1,v1)

xlabel 'Time (s)'

ylabel 'Velocity (m/s)'

title 'Graph of Time vs Velocity'

figure()

plot(t1,h1)

xlabel 'Time (s)'

ylabel 'Height (m)'

title 'Graph of Time vs Height'

In parachute task 2, a skydiver starts at 4000 m and descends for 60 seconds at which time his chute opens instantaneously. In task 3, however, we will incorporate chute deployment which occurs over a few seconds. During chute deployment, the effects of the drastic changes in chute area can be approximated using the polynomial equation shown below. Copy your script from task 2 to begin task 3 and follow the directions below. (Note: in step 3, you are asked to record the acceleration values experienced during the jump.)

Explanation / Answer

There are three parts to this project,First ,finding the time at which the main parachute should be deployed.second, calculating the landing position, and third, determining the "jerk force". Allowing discontinous force on the skydriver, a common model for his / her vertical motion is the differential equation

v =g - { k1v2 before deployment

{k2v after deployment

k1,k2 may be determined by using the terminal velocitybefore deployment and assuming the landing speed in the terminal velocity after deployment.

To find the time at which the main parachute must be deployed, we find the distance that the skydriver falls if his/her main parachute fails to deploy and he/she then pulls his/her reserve, and require the terminal velocity for the reserve parachute be attained before impact with the ground.

The distance fallen is the integral of the velocity(area under the graph of v against t).

By numerically solving v = g-k1v2, v(0)=0, where k1=0.001

Here,

Consider the above question,

v = (g-(5/3)*u)*ts =0.315

height = height -h

h = u*ts

= 0*0.1 = 0

therfore height = 4000 - 0 = 4000

u = u+v

= 0 + 0.315 = 0.315

t = t+ts = 0.1

%plotting

"Graph of time vs velocity

I

velocity(m/s) I I----------------I

0.315 I I I I

I I I I

I-----I-----I-----------I-----------------------------------------

0 0.1 time(s)

Graph of time vs height

4000 I I

I I

height(m) I I

I I

I--------I---------------------------------------

0 0.1 time (s)

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