Here is the question I'm dealing with:
A ball with mass 0.15 kg is thrown upward with initial velocity 20 m/s from the roof of a building 30 m high. There is a force due to air resistance of magnitude |v|/30 directed opposite to the velocity, where the velocity v is measured in m/s.
Find the time that the ball hits the ground.
NOTE: I have calculated the highest height the ball reaches: it is 45.783 m
m dV/dt = -mg + V/30
dV/dt = -g + V/30m
dV/dt = (-30mg + V)/30m
dV/(-30mg + V) = dt/30m
ln(-30mg + V) = t/30m + C
e^(ln(-30mg + V)) = e^(t/30m + C)
-30mg + V = e ^(t/30m + C)
V = Ce^(t/30m) + 30mg
*Since the projectile is instantaneously at rest, V(0) = 0
Therefore, C = -30mg
So:
V = 30mg - 30mg*e^(t/30m)
Now to integrate to get our position function:
X = 30mg*t - (30m)^2*g*e^(t/30m) + C
Since X at t = 0 is 45.783 m, C is equal to (30m)^2*g
I know that setting X= 0 will tell me the time at which the projectile hits the ground, but I cannot for the life of me figure out how to arrive at an answer for t without just going to Wolfram Alpha.
Is there a better way for me to be attempting what I'm attempting?
I apologize for not having proper formatting for my work.
Thanks!
The equation for $v$ (see my comment above) is $$m\dot v=-mg-kv$$Let $v=\dot y$.
Then$$y=y_0-\frac{mg}kt+\frac mk\left(v_0+\frac{mg}k\right)\left(1-e^{-\frac kmt}\right)$$Given $y_0=30$ and $v_0=20$, one solves (by a numerical method) $$y=0$$with rspect to $t$ and finds $t=5.1285 s$ .