Question

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

If you assume drag force is proportional to speed, then
dragforce= k*speed
and at terminal speed, dragforce=weight.
80kg*g=k*40m/s
so you can calculate k.
Then, calculate the drag force at 30.

wait, where did you get the 40 m/s?
and i still don't understand how to find the acceleration.
and if i calculate k that way..then wouldn't the value of k change when her speed changes?

40 was a typo, should have been 50.

vfinal^2 -vinitial^2= 2*acceleration*distance

but distance= average velocity *time
= (Vfinal + Vinitial )/2

solve for acceleration.

k is a constant, it wont change with speed if you assume linear proportionality.

Answer 1

Given m = 80.0 kg, vT = 50.0 m/s, we write
mg = DρAvT
2
2
which gives
DρA
2
= mg
vT
2 = 0.314 kg m
(a) At v = 30.0 m/s,
a = g − DρAv2 2
m
= 9.80 m/s2 −
(0.314 kg/m)(30.0 m/s)2
80.0 kg
= 6.27 m/s2 downward
(b) At v = 50.0 m/s, terminal velocity has been reached.
Fy Σ = 0 = mg − R
⇒R = mg = (80.0 kg)(9.80 m s2 ) = 784 N directed up
(c) At v = 30.0 m/s,
DρAv2
2
= (0.314 kg/m)(30.0 m/s)2 = 283 N upward