A skydiver with a mass of 80 kg (including parachute) jumps out of an airplane a

Question

A skydiver with a mass of 80 kg (including parachute) jumps out of an airplane at an altitude of 4000 in. the initial vertical velocity is 0 m/s. He free-falls for 40 seconds before opening a parachute. The parachute has a 10 m diameter when completely opened (you can assume it opens instantaneously). Create a program in MATLAB that calculates and plots the skydiver?s vertical speed and altitude as a function of time using the following equations and values. Remember that the skydiver cannot exceed the terminal velocity and the parachute has a failsafe to open at 1000 m. Hints: There are different terminal velocities with and without the parachute open, assume that the parachute opens instantaneously Constant values Calculation of terminal velocity Calculation of downward acceleration (derived from Sigma F = m * a) Kinematics Equations setup for a loop Setup the diameter of the parachute. initial altitude, time of opening the parachute, and the time step as user inputs. In addition to the plots, output the answers to the following questions the command window 1. Ls terminal velocity reached before the skydiver opens the parachute? If so, at what time did the skydiver reach terminal velocity? 2. At what altitude did the parachute open? 3. Does the skydiver open the parachute or did the failsafe mechanism open it?

Explanation / Answer

(1)First of all we will apply the first equation of motion i.e. Vi = Vi-1 +at where Uf is final velocity after 40 seconds and Ui is initial velocity i.e. Zero. and a is acceleration and t is time i.e. 40 seconds.
acceleration will be calculated by the above given equation
when we put this equation in above equation a quadratic equation get formed in Vi hence we will able to calcualate the value of Vi.
For calculation of terminal velocity we know that terminal velocity is the velocity attained by the free fall object when buyoynacy and drag force equate the weight of object
Drag force is equal to FD = (1/2) Cd p AU2 and buoyancy force is FB = pgV where V is volume
When we equate the sum of buoyancy and Drag force i.e. FD +FB = W --(1) where W is weight of object i.e. (80*9.81)
Therefore we can calculate the value of terminal velocity (U) as we know all the values.
Here we apply all the values of before opening of parachute.
Once we find out the U (terminal velocity) from the above equation we will come to know that whether terminal velocity is achieved or not.
(2) As it is mentioned that parachute was open after 40 seconds so we will calcualte the distance covered by the object in 40 second with the help of second eqaution of motion i.e

Hence we will calcualte the distance covered in 40 seconds.
(3) If the above calcualted distance is less than 3000 then skydriver will open the parachute.
If the value of above calculated distance is more than 3000 then failsafe mechanism will open the parachute.

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