A 54kg skydiver jumps out of an airplane. We assume that the forces acting on th

Question

A 54kg skydiver jumps out of an airplane. We assume that the forces acting on the body are the force of gravity and a retarding force of air resistance with direction opposite to the direction of motion and with magnitude cv^2 where c=0.14kg/m and v(t) is the velocity of the skydiver at time t (and upward is positive velocity). The gravitational constant is g=9.8m/s2.

a. Find a differential equation for the velocity vv :

b. Determine the terminal velocity in meters per second for free-fall (no parachute).

Explanation / Answer

Given : m= 54 kg, c= 0.14 kg/m, g = 9.8 m/s2

Solution :

(a) The velocity (vt) of a skydiver falling to the ground is given by:

m (dvt/dt)= mg- cv2

Re arranging

dvt/dt= g- (c/m)v2

Form of differential equation would be:

[1/g-(c/m)v2] dvt/dt = 1

(b)

Terminal velpcity can be find out as :

vt= [2mg/cm]1/2

= (2g/c)1/2

= [(2×9.8)/0.14]1/2

= 11.83 m/s

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