A 72.5-kg skydiver reaches a terminalspeed of 56.2 m/s with her parachuteundeplo

Question

A 72.5-kg skydiver reaches a terminalspeed of 56.2 m/s with her parachuteundeployed. Suppose the drag force acting on her is proportional tothe speed squared, or Fdrag =kv2. (a) What is the constant of proportionalityk? (Assume the gravitational acceleration is 9.8m/s2.)
1

(b) What was the magnitude of the acceleration when she was fallingat half terminal speed?
2 m/s2 (a) What is the constant of proportionalityk? (Assume the gravitational acceleration is 9.8m/s2.)
1

(b) What was the magnitude of the acceleration when she was fallingat half terminal speed?
2 m/s2

Explanation / Answer

Therefore mg = kv2
then k =(72.5kg)(9.8m/s2)/(56.2m/s)2
     = 0.225
(b) At half the terminal speed we have mg -k(v/2)2 = ma Then a = g - kv2/4m
      = 9.8m/s2 -(0.225)(56.2m/s)2/(4*72.5kg)       = 9.8m/s2 - 2.45m/s2       = 7.35m/s2

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