terminal velocity (M1)

Jun 2016
13
0
ontario
A skydiver of mass m kg steps out of an aircraft and starts to fall. The air resistance when his speed is v ms^-1 is given by kv^2 newtons. If his terminal speed is 40 ms^-1, find an expression for k in terms of m. (I did this part, got (m/160), which is correct)

When his speed reaches 16 ms^-1 he adopts a position for which the air resistance is 4kv^2 newtons. Find his new terminal speed (Did this as well, got 20m/s which is also correct).

Find also the acceleration with which he is falling (need help with both of these):

(a) just before he adopts the new position. (answer in book is 8.4m/s/s)

(b) just after he adopts the new position. (answer in book is 3.6m/s/s)

I have no idea how to go about solving (a) and (b). please help!
(Happy)
 

romsek

MHF Helper
Nov 2013
6,738
3,034
California
Right before the new position the diver's velocity is $v=16 m/s$

so the net force downward is given by (assuming as you seem to that $g=10$)

$m g - k v^2 = m g - \dfrac {m}{160}{16^2} = m\left(10-\dfrac{256}{160}\right) = 8.4m~~N$

and thus the net acceleration downward is $\dfrac F m = 8.4 ~~m/s^2$

Right after adopting the new position his air resistance is $4k = \dfrac {m}{40}$

His velocity is as yet unchanged and so $v = 16 m/s$

So the net force is $F = m g - \dfrac {m}{40}(16^2) = m\left(10 - \dfrac{256}{40}\right) = 3.6m ~~N$

and thus the net acceleration downward is $\dfrac F m = 3.6~~ m/s^2$
 
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