Question

A small stone of mass m, is thrown vertically upward with initial speed ,u. If the air resistance at speed v is mkv2, where k is a positive constant. Show that the stone returns to its starting point with speed given by v=u/((1+(ku2/g))1/2)

Answer 1

$$W=-mg$$

$$F_r=-mkv^2$$

$$ma=m\frac{dv}{dt}=-m(g+kv^2)$$

$$v\frac{dv}{dx}=-(g+kv^2)$$

$$-2\int_0^0dx=\int_u^v\frac{dv^2}{g+kv^2}$$

$$v=u\left(1+\frac{ku^2}{g}\right)^{-\frac{1}{2}}$$