Answer to Question #101479 in Classical Mechanics for israel

Question

A small stone of mass, is thrown vertically upward with initial speed ,u. If the air resistance at speed is mkv , where v is a positive constant. Show that the stone returns to its starting point with speed given by v=u/(( 1+ku .

Answer 1

As per the question,

Resistivity force $F_r=-mkv$

And force due to gravity $F_g=-mg$

So, net force $ma= F_g+F_r$

$\Rightarrow ma=-(mg+mkv)$

we know that $a=\dfrac{dv}{dt}=\dfrac{dv}{dt}\times \dfrac{dx}{dx}=v\dfrac{dv}{dx}$

$\Rightarrow ma=-m(g+kv)$

$\Rightarrow v\dfrac{dv}{dx}=-(g+kv)$

$\Rightarrow \int_u^v \dfrac{vdv}{g+kv}=\int_0^x dx$

At the final condition, net displacement =0

(There is mistypes data in the question, either $mkv^2$ or the answer given will be something else)