A body of mass m is thrown straight up with velocity `v_0`. Find the velocity `v^'` with which the body comes down if the air drag equals `kv^3`, where k is a constant and v is the velocity of the body.

Explanation

While going upward, from Newton's second law in vertical direction:
`m(vdv)/(ds)-(mg+kv^2)` or `(vdv)/((g+(kv^2)/(m)))=-ds`
At the maximum height h, the speed `v=0`, so
`underset(v_0)overset(0)int(vdv)/(g+(kv^2//m))=-underset(0)overset(h)intds`
Integrating and solving, we get
`h=(m)/(2k)1n(1+(kv_0^2)/(mg))` (1)
When the body falls downward, the net force acting on the body in downward direction equals `(mg-kv^2)`,
Hence net acceleration, in downward direction, according to second law of motion
`(vdv)/(ds)=g-(kv^2)/(m)` or, `(vdv)/(g-(kv^2)/(m))=ds`
Thus `underset(0)overset(v^')int(vdv)/(g-kv^2//m)=underset(0)overset(h)intds`
Integrating and putting the value of h from (1), we get,
`v^'=v_0//sqrt(1+kv_0^2//mg)`