Question
Velocity of a free-falling parachutist. Please show working, Thanks.
The velocity of a free-falling parachutist, subject to linear drag, can be calculated as the following function of time: V(t) = gm/c(1-e-(c/m)t) where v is the velocity (m/s), g is the gravitational constant of 9.8m/s2, m is the mass of the parachutist equal to 68.1 kg, and c is the drag coefficient of 12.5kg/s. We wish to know how far the parachutist has fallen after a certain time t. This distance is provided by y = v(t)dt Where y is the distance in m. Substituting the velocity function in this equation, we have y = gm/c (1-e-(c/m)t)dt Analytically integrate the above distance function with the initial condition that y = 0 at t = 0 .Evaluate the result at t = 10. Evaluate (on paper) the above integral using the multiple-application of the trapezoidal rule. You need to experiment with different values for segment, in order to achieve the best possible answer. Provide the related percent relative error against the real analytical solution. Evaluate (on paper) the above integral using the multiple-application of Simpson's 1/3 rule. You need to experiment with different values for segment, in order to achieve the best possible answer. Provide the related percent relative error against the real analytical solution.
Explanation / Answer
the velocity of the freely falling object is
v = u + gt
where u is initial velocity of object,g = 9.8 m/s^2 and t is time
the distance covered by the object in time t is
S = ut + (1/2)gt^2
when u = 0
S = (1/2)gt^2
and v = gt
