Saved Bookmarks
| 1. |
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. |
|
Answer» Solution :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)` |
|