Physics

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.

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asked by Jean
  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

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    posted by amna

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