A rain drop with radus 1.5 mm falls from a cloud at a height 1200 m from ground . The density of water is 1000 `kg//m^(3)` and density of air is `1.2 kg//m^(3)`. Assume the drop was spherical throughout the fall and there is no air drag. The impact speed of the drop will be :

Explanation

The drop reaches a terminal speed `v_(t)` when the gravitational force on it is balanced by the air drag force on it, so its acceleration is zero. We could then apply Newton.s second law and the drag force equation to find `v_(t)`, but Eq. 6-30 does all that for us.
Calculations: To use Eq. 6-30, we need the drop.s effective cross-sectional area A and the magnitude `F_(g)` of the gravitational force. Because the drop is spherical, A is the area of a circle `(piR^(2))` that has the same radius as the sphere. To find `F_(g)`, we use three facts: (1) `F_(g)=mg`, where m is the drop.s mass, (2) the (spherical) drop.s volume is `V=4//3piR^(3)` and (3) the density of the water in the drop is the mass per volume, or `rho_(w)=m//V`. Thus, we find
`F_(g)=Vrho_(w)g=(4)/(3)piR^(3)rho_(w)g`.
We next substitute this, the expression for A, and the given data into Eq. 6-30. Being careful to distinguish between the air density `rho_(a)`, and the water density `rho_(w)`, we obtain
`v_(t)=sqrt((2F_(g))/(Crho_(a)))=sqrt((8piR^(3)rho_(w)g)/(3Crho_(a)piR^(2)))=sqrt((8Rrho_(w)g)/(3Crho_(a)))`
`=sqrt(((8)(1.5xx10^(-3)m)(1000kg//m^(3))(9.8m//s^(2)))/((3)(0.60)(1.2k//m^(3))))`
`=7.4m//s~~27km//h`
Note that the height of the cloud does not enter into the calculation.
(b) What would be the drop.s speed just before impact if there were no drag force?