A skydiver of mass 80.0 kg jumps from a slow-moving aircraft and reaches a termi

Question

A skydiver of mass 80.0 kg jumps from a slow-moving aircraft and reaches a terminal speed of 50.0 m/s. (a) What is her acceleration when her speed is 30.0 m/s? What is the drag force on the skydiver when her speed is (b) 50.0 m/s and (c) 30.0 m/s?

Am I on the right track?? Please help me with the details.
So if at terminal velocity the skydiver is no longer accelerating the equation of forces used should be:

mg = bv (b being a constant)
I am also not quite sure if I should use 9.8 m/s2 or if it is ok to round and use 10m/s2 for physics.

Given: m = 80 kg, vt = 50 m/s.

(a) What is the acceleration of the skydiver, when her speed is 30 m/s?
(b) What is the drag force on the diver when her speed is 50 m/s?
(c) What is the drag force on the diver when her speed is 30 m/s?

a) Using we can find DA.  DA = 2mg/vt2 = 0.6272 kg/m.
Therefore a = dv/dt = 9.8 m/s2 - (0.6272/(160 m))v2.
When v = 30 m/s, a = 6.272 m/s2.

b).When v = vt = 50 m/s, the drag force is equal in magnitude to mg = 784 N.

c.) When v = 30 m/s, the total force is equal in magnitude to 
ma = (80 kg)(6.272 m/s2) = 502 N.  The magnitude of the drag force is mg - ma = 282 N.

Explanation / Answer

yeah!! exactly that is how its done. after a maximum velocity the acceleration stops due to high drag force. and in most of the problems g is taken to be 10m/sec2.

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