Why Work done = mgh.

wen a raindrop falls
force acting on it is mg(that is f)
and the displacement in the direction of the force(that is s cos(theta)) is h

so work done=f s cos(theta)=mgh


also
by conservation of energy
work+change in potential energy=0
so work done=-(change in potential energy)

Work done, W = F x D, .......... when, theta = 0.

Force, F = mass x the gravity acceleration = mg

diatnce, D = height, h,
Thus,
W = mgh = 7.2x10^-9 x 9.8x1.8 = 3.2x10^-7 >======< ANSWER

You seem to have an issue with the term "work done". Let me just clarify by saying that "work done" refers to the transfer of energy.

In your case, provided that there is no air resistance (otherwise it'll be different) this is how you'd do this problem:

Initial potential energy = mgh = 7.2x10^-9 * 9.8 * 4.5 = 3.24x10^-07 (the answer you gave used g=10, but the principle remains the same.)

Because there was no air friction, and it is moving at a constant velocity (net force = 0) ALL the potential energy will go into the kinetic energy (if there was any form of external forces, this would NOT be the case, the Ep would dissipate into other forms of energy). You might get a question where it gave you that info, but, for example, also gave you the kinetic energy at the end or something. In that case, the "work done" would be Ep - Ek (potential energy, mgh, - kinetic energy, 1/2mv^2)

Hope this clears it up.

W = F x d or W = F x d x cos (theta) if your considering a force at some angle relative to the direction of travel.

In addition to what was said in the above post the initial force on the rain drop must have been the force due to gravity which is equal to mass x acceleration due to gravity (9.81 m/s^2).

Since the force is parallel to the direction of travel W = is simply F x d. If you use F x d x cos(theta) theta in this case theta would be 0 since the force makes no angle with the direction of travel (they are in the same direction). This means cos (0) = 1 so we get W = F x d