# terminal velocity (M1)

#### wings

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)

(Happy)

#### romsek

MHF Helper
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 $\dfrac F m = 3.6~~ m/s^2$

2 people