# Physics problem - about a shell. Differential equation

Can you help me please, to write the differential equation for this problem, and give me an idea how to solve this equation.

A shell of mass $2$ kg is shot upward with an initial velocity of $200$ m/sec. The magnitude of the force on the shell due to air resistance is $|v| =20$. When will the shell reach its maximum height above the ground? What is the maximum height?

I wrote: $$\begin{cases}2\dot{v}=\frac{v}{20}-2g\\v(0)=200\end{cases}$$

But I don't know if it ok.

thanks :)

• Both gravity and air resistance are working against the mass on the way up. The sign on the resistance term should be reversed. Commented Jan 11, 2014 at 8:23
• so you say: $2\dot{v}=-\frac{v}{20}-2g$?thanks
– Iuli
Commented Jan 11, 2014 at 8:24
• Yes, otherwise if the initial speed was enough, the acceleration would be positive and increasing. Commented Jan 11, 2014 at 8:26

The equation: $$\frac{dv}{dt}+\frac{v}{40}=-g$$ is a first-order linear ODE with initial condition $v(0)=200$. The solution (via integration factor) is: $$v(t)=e^{-t/40}[200+40g]-40g.$$ At the maximum height the object's velocity will be zero (at the instant that it starts to come back down). Thus set $v=0$ and solve for $t$ to get $t=-40\ln{\frac{40g}{200+40g}}=16.49$ seconds. To determine at what height, $h$, this occurs, realize that $v=\frac{dh}{dt}$ and integrate the expression $$v=\frac{dh}{dt}=e^{-t/40}[200+40g]-40g$$ with the initial condition of $h(0)=0$ to get $$h(t)=40(200)+40^{2}g-40gt-e^{-t/40}[40(200)+40^{2}g].$$ Now plug $t=16.49$ into this expression to find $h=1536$ meters.