Question

Hi, I need help with this question.

Assume that the upward force experienced by a bungee jumper is given by a linear relationship Fu = -c'v, where c' is a first order drag coefficient, v is the velocity of the bunjee jumper, and Fu is the force acting in the upward direction. Using the concepts and relations discussed in the course for this problem, answer the following:

Assume that there are two jumpers. Assume that the first jumper is 85 kg and that the second jumper is 60 kg. Calculate the drag coefficients for each of these jumpers if they must both have the same velocity of 20 m/s in 6 seconds after jumping. Both jumpers are at rest initially and both jump from the same level (elevation).

Answer 1

First Jumper (85kg) Second Jumpor (60 kg) direction Net = F = force acting on jumper in dovencwars Geranitational Force - Drag force mg - co famiga m, dr = m.g. cv, dt myckz m. du = m. g - 2 due = at (g- del m) = at du, lig- culm) Integrating on both sides en (g- civolom, 2 = t tz (-co/m) en (gaul m) = t + (-calon)

@t-o uzo @t- ow.co + z = - m. ing cy: - m lng. en in sy = -cita en 1-civa 7 = út - meg @ t-orec v = 2 = 20 m/sec =

mfg-cs) = ++ and you (-cilm) m (1- Síxi 't m.=85;t:6; =20 In 1-0.023945129216) = -0.0705 4823 st c=0 OR 39.0431 | + | Du |- c , 1 : - GV men me = = 0 OR 27. 5598 C' = 39.0431) G' = 27.5598]