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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 :)

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

1 Answer 1

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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.

Hope this helps.

Cheers,

Paul Safier

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