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Question: After a free falling sky diver of mass $m$ reaches a constant velocity of $v_1$, her parachute opens, and the resulting air resistance force has strength $Kv$. Once the parachute opens, the equation of motion is

$$ \frac {dv}{dt} = g - Bv $$

where $ B = \frac Km $, $K$ is the constant of proportionality and $g$ denotes the value of gravitational acceleration.

Show that the equation for the speed of the sky diver $t$ seconds after the parachute opens is

$$ v = \frac{mg}{K} \left[ {1} + ({\frac{K}{mg}v_1 - 1)e^{-{Kt\over m}}} \right] = 0$$

What I have done:

$$ \frac{dv}{dt} = g-Bv $$

$$ \int \frac{1}{g-Bv} dv = \int dt $$

$$ -\ln|g-Bv| \cdot \frac{1}{B} = t + Bc$$

$$ g-Bv = Ae^{-Bt}$$

$$ g-Ae^{-Bt} = Bv $$

$$ {g-Ae^{-Bt}\over B} = v$$

When $t$ $=$ $0$ , $v$ $=$ $v_1$

$$ {g-A\over B} = v_1$$

$$ A = g-Bv_1 $$

$$ v = {g-Ae^{-Bt}\over B} $$

$$ v= {g-(g-Bv_1)e^{-Bt}\over B} $$

$$ v= {g+(Bv_1-g)e^{-Bt}\over B} $$

But $ B = \frac Km $

$$ v= {mg+({K\over m}v_1-g)e^{-({K\over m})t}\over K} $$

Now am I stuck...

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  • $\begingroup$ In the last expression you obtained, the $m$ needs to multiply the bracketed term as well, then you have the result you want. $\endgroup$ – David Quinn Jan 29 '16 at 21:09
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From $$v={g+(Bv_1-g)e^{-Bt}\over B}$$ and using $$B=\frac{K}m$$ we get $$v={g+(\frac{K}mv_1-g)e^{-\frac{K}mt}\over \frac{K}m}$$ Multiplying top and bottom by $m$ gives $$v={m(g+(\frac{K}mv_1-g)e^{-\frac{K}mt})\over K}$$ Multiply the $m$ through the bracket in the numerator $$v={mg+(Kv_1-mg)e^{-\frac{K}mt})\over K}$$ and remove the 'common' factor of $mg$ $$v={mg(1+(\frac{K}{mg}v_1-1)e^{-\frac{K}mt})\over K}$$ and identify $K$ with $mg$ $$v=\frac{mg}K\left[1+(\frac{K}{mg}v_1-1)e^{-\frac{K}mt}\right]$$

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