Differential Equation -how to solve for the position of the rock on the given problem. A $2 kg$ rock is thrown up from a $180 m $ high cliff with an initial velocity of $20 m/s$. Assume the force due to air resistance is equal to $0.5 v$, where $v$ is the instantaneous velocity, determine the position of the rock as function of time.
-Assume $x$ is the altitude of the rock and $v$ is its velocity. Assign a positive sign for upward direction.
I've came up with this formulas but i'm not sure if its correct:
solving for $v(t)$:
 $$ \frac{dv}{dt} = -9.8 $$ where, $c=20$
solving for $p(t)$:
 $$\frac{dp}{dt}=-9.8t-20$$
I don't understand the condition stated.
 A: Notice, 
The motion is under Earth's gravity & the acceleration due to gravity $g$ does not depend on the mass of the body.
Now, let the altitude be $x$ at any time $t$ from measured from the ground. Since the motion is in the downward direction, hence net acceleration in the down ward direction $$\frac{dv}{dt}=g-\frac{0.5v}{2}=\frac{4g-v}{4}$$
  $$\frac{4dv}{4g-v}=dt$$ $$\int \frac{4dv}{v-4g}=-\int dt$$ $$4\ln(v-4g)=-t+C_1$$
 Now, setting the value of the velocity $v=20$ at time $t=0$, we get $$4\ln(20-4g)=0+C_1\iff C_1=4\ln(20-4g)$$ $$4\ln(v-4g)=-t+4\ln(20-4g) \iff 4\ln\left(\frac{v-4g}{20-4g}\right)=-t\iff \ln\left(\frac{v-4g}{20-4g}\right)=-\frac{t}{4}$$  $$\frac{v-4g}{20-4g}=e^{-\frac{t}{4}}$$$$ v=(20-4g)e^{-\frac{t}{4}}+4g$$
Since, the velocity is downwards increasing with decrease in the altitude $x$ measured from the ground, hence $v=-\frac{dx}{dt}$, we get $$-\frac{dx}{dt}=(20-4g)e^{-\frac{t}{4}}+4g$$
$$\int dx=-(20-4g)\int e^{-\frac{t}{4}}dt-4g\int dt$$ $$x=-(20-4g)(-4)e^{-\frac{t}{4}}-4gt+C_2$$$$x=16(5-g)e^{-\frac{t}{4}}-4gt+C_2$$
Now, at time $t=0$, altitude $x=180$
$$180=16(5-g)e^{0}-0+C_2\iff C_2=180-80+16g=100+16g$$
$$x=16(5-g)e^{-\frac{t}{4}}-4gt+100+16g$$  setting the value of $g=9.8$
$$x=16(5-9.8)e^{-\frac{t}{4}}-4(9.8)t+100+16(9.8)$$ 
$$x=-76.8 e^{-0.25 t}-39.2t+256.8$$
Hence, we get 
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{Position/altitude (x) of rock as a function of time (t):}}\ \color{blue}{x=256.8-76.8 e^{-0.25 t}-39.2t }}$$
A: You forgot the force due to air resistance. So the differential equation should be $$\frac{dv}{dt}=-9.8+0.5v$$ with the initial condition $v(0)=-20$ If the rock is thrown downwoard!
I let you solve it from there.
