A 60.4-kg skydiver reaches a terminal speed of 53.9 m/s with her parachute undep

Question

A 60.4-kg skydiver reaches a terminal speed of 53.9 m/s with her parachute undeployed. Suppose the drag force acting on her is proportional to the speed squared, or Fdrag = kv2. (a) What is the constant of proportionality k? (Assume the gravitational acceleration is 9.8 m/s2.)
1

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

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

Explanation / Answer

at terminal velocity,

Fdrag = mg

k x 53.92 = 60.4 x 9.81

k = 0.204

b) half of therminal velocity,

Fdrag = 0.204 x (53.9/2)2 = 148.16

a = mg - Fdrag / m = 444.364/60.4 = 7.36 m/s2

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