A body of mass M is dropped from a height h on a sand floor. If the body penetrates x cm into the sand, the average resistance offered by the sand to the body is:

If the body strikes the sand floor with a velocity V, then mgh = $\frac{1}{2}{\mathrm{mV}}^2$

With this velocity V, when body passes through the sand floor it comes to rest after traveling a distance x. Let F be the resisting force acting on the body. Net force in downward direction = Mg-F

   We know that work done by all the forces is equal to change in KE: (Mg-F)x-0-$\frac{1}{2}{\mathrm{MV}}^2$

or   (Mg-F)x = -Mgh or Fx = Mgh + Mgx

or F = Mg(1+$\frac{\mathrm{h}}{\mathrm{x}}$)