Force exerted to a falling object

1. The problem statement, all variables and given/known data

A rock with a weight of 120 N free falls from height of 2.00m collides with with earth and digs 60 mm into it. What is the Earth's average force of resistance


2. Relevant equations

F=m*g => m=F/g

m*g*h = 1/2*m*v^2

v=√(2*g*h)

3. The attempt at a solution

I found the velocity of a rock just before collision using the law of conservation of energy. I now know the velocity which I got 6.3 m/s but now im stumped on how do i find the forces exerted by earth to stop the rock from digging deeper.

 

i assume you substitute s with 0.006 m right. I tried that in that case

[itex]v^{2}_{f}[/itex]=[itex]v^{2}_{i}[/itex] + 2 * a * x

0=6.3^{2}+2*a*0.06
0=39.69+0.12a
-0.012a=39.69 / (-0.12a)
a=330.75 m/s^2

F=m*a

F= 12 * 330.75 = 3969 N

however the answer should be 4.1 kN

 

getting closer :P thanks I guess it was just an error in rounding the number