Question

"an object is dropped from a great height. its acceleration due to gravity is 9.8 meters per second per second and its deceleration due to air resistance is proportional to its velocity. after a long period of time, its terminal velocity of 30 meters per second is achieved. what was its velocity after 10 seconds?"

Answer 1

When the object reaches its terminal velocity, its net acceleration is zero:
$a_{net}=0$
The net acceleration on the object is given by the difference between the gravitational acceleration g (directed downward) and the acceleration due to the air resistance, $a_r$:
$a_{net} = g - a_r$ (1)
the acceleration due to the air resistance is proportional to the velocity, so we can write it as the product between a certain coefficient k and the velocity:
$a_r = kv$ (2)

When the terminal velocity is reached, the net acceleration must be zero. We can rewrite (1) using (2):
$a_{net} = g-kv=0$
from which we find k:
$k= \frac{g}{v}= \frac{9.8 m/s^2}{30 m/s}=0.33 s^{-1}$

Then we know the object moves by accelerated motion, with net acceleration $a_{net}$, so its velocity at time t is given by:
$v = a_{net} t = (g-kv)t$
And by re-arranging this equation and using t=10 s and $k=0.33 s^{-1}$, we find the velocity of the object after 10 seconds:
$v= \frac{gt}{1+kt}= \frac{(9.8 m/s^2)(10.0s)}{1+(0.33 s^{-1})(10.0 s)}=22.8 m/s$