A 58.3-kg skydiver reaches a terminalspeed of 52.2 m/s with her parachuteundeplo

Question

A 58.3-kg skydiver reaches a terminalspeed of 52.2 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.)
Enter anumber.

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

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

Explanation / Answer

The drag force = kv2 We have Fdrag - mg = 0            kv2 = mg              k = mg /v2                 = 58.3 * 9.8 / 52.2*52.2                 = 0.209 b)           Fdrag - mg =ma          a =(kv2 - mg ) / m            = (0.209*52.22 - 58.3*9.8) / 58.3            = .......... m/s2 Solve it.

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