A skydiver of mass 75.0 kg jumps from a slow-moving aircraft and reaches a terminal speed...

A skydiver of mass 75.0 kg jumps from a slow-moving aircraft and reaches a terminal speed of 47.1 m/s.

(a) What is her acceleration when her speed is 30.0 m/s?

magnitude	m/s²
direction	---Direction--- upward downward

(b) What is the drag force on the skydiver when her speed is 47.1 m/s?

magnitude	N
direction	---Direction--- upward downward

(c) What is the drag force on the skydiver when her speed is 30.0 m/s?

magnitude	N
direction	---Direction--- upward downward

Expert Solution

Given mass 75 kg, terminal velocity 47.1 m/s

   the drag force = the gravity force when terminal velocity is reached
           = 75*9.8 = 735 N
b)       F(drag) = k . V^2 ==> k = 735/47.1^2 = 0.3313

c) Fdrag at 30 m/s is F= kv^2 = 0.3313*900 = 298.18 N

a) acceleration is a= (735 - 298.18) / 75
= 5.824 m/s2

genius_generous answered 2 years ago

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