A problem related to Work done by falling bodies : Expert's attention much needed! I'm having a lot of trouble with this question, that I've found in my textbook. I've solved it in my own way and it's very simple! 
But the solution in the book is totally different. It doesn't make any sense to me and the answers don't match either!! I think my method & answer are correct, but I'm not getting any strong support or proof, that's why I came here and posted... 
So, the problem goes as follows -

  
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*A hammerhead weighing about 1 kg hits on a nail at 0.8 m/s. the nail penetrates 0.02m deep in the ground. Calculate the average Resisting Force of the ground.

So I did it in this way- 
$$ \frac {mv^2}{2} = Fd .$$ 
here d = 0.02m
(As we know, Work Done = total energy, [important]  the whole potential energy was transferred into Kinetic energy just before the nail was hit, & the energy was 0 when the nail stopped)
So, I got $F$ from here! 
But the method given in the book is like this :
First they have calculated the height from where the hammerhead fell with the help of this law -
$$ v^2 = 2gh $$
then they have said that, [important] as the hammerhead penetrated into the ground 0.02m, so it has traveled  (h+0.02)m and we need to calculate the Potential Energy using this height.
(Isn't this insane!!?)
Then the book says : total potential energy 
$$ mg(h+0.02) = Fd, $$
& they have found the $F$ like this. 
So, is my method wrong!? What is wrong in there, because two answers don't match! Why didn't they use $W = mgh$, rather than $mg(h+0.02)$ ??
My logic - we use $W = mgh$ law as long as $g$ is constant, but in the book they used - 
$$ mg(h+0.02) = (mgh + mg*0.02). $$ 
[ Here's the thing,  how come you to use $mg*0.02$  when the acceleration wasn't even $g$ when the nail traveled the 0.02m, since after hitting the nail the hammerhead will not travel the 0.02m with the acceleration $g$ ? It rather will experience some retardation, otherwise the hammerhead & the nail will keep goin' in forever! ]
If my method is 100% correct then how do I prove them wrong? Please tell me some mathematical way, because my logic isn't enough for some of my friends! :-(
thanks for reading this whole big question.
any helps will be appreciated greatly! and sorry for my poor english! 
 A: You have assumed that the entire energy dissipated by friction is the KE of the hammer on impact.  But the problem details that besides the energy on impact, the hammer gains energy by dropping further.  It loses PE corresponding to dropping an additional $0.02m$.  That energy has to go somewhere, and it goes into work done against friction.
So you could revise your formula to be:
$$\frac{mv^2}{2} + mg(0.02m) = Fd$$

...after hitting the nail the hammerhead WILL NOT travel 0.02m at 'g'...

That's not what the equation is saying.  It's just using the formula for the difference in potential energy.
$$\Delta PE = mg\Delta h$$
The $g$ there isn't describing the acceleration of the hammer during the impact.  It's using the strength of the gravitational field to calculate the energy released by decending the additional $0.02m$ during the impact.
A: They are right, and your logic is also right but you don't apply it correctly.
What stops the hammerhead? A bigger acceleration than $g$ but in opposite direction. In short, there are two forces here, each one imposing its acceleration, but while the gravitation acts until the hammerhead stops, the force $F$ acts only along $d$.
The hammerhead falls along a distance $h + 0.02m$, as they say, and with the acceleration $g$. From this you calculate the mechanic work invested by the hammerhead.
Now, as I said, the hammerhead is gradually stopped due the bigger acceleration in opposite direction, imposed by the ground resistance $F$. 
Which work does this force? $F \cdot d$. The work done by the ground resistance is equal in absolute value with the total work of the hammerhead, but opposite in direction. So, indeed their formula is correct.
Simple!
A: Total energy (KE + PE) of hammerhead just before impact:
$$E_i = \frac{1}{2} 1 \mathrm{kg} \left( 0.8\mathrm{\frac{m}{s}} \right)^2 + U_0$$
Total energy after impact:
$$E_f = U_0 - 1\mathrm{kg}\cdot g \cdot 0.02\mathrm m$$
Work done by hammerhead is
$$W = E_i - E_f =  \frac{1}{2} 1 \mathrm{kg} \left( 0.8\mathrm{\frac{m}{s}} \right)^2 + 1\mathrm{kg}\cdot g \cdot 0.02\mathrm m$$
The average force is then
$$\bar F = \frac{W}{0.02\mathrm m}$$
But, assuming the hammer fell from rest a distance $h$ in order to have the stipulated KE just before impact, it is true that
$$\left( 0.8\mathrm{\frac{m}{s}} \right)^2 = 2gh$$
Thus
$$W = 1\mathrm{kg}\cdot g\cdot \left(h + 0.02\mathrm m \right) $$
