We can solve the problem by using the law of conservation of energy.
Initially, the rock has only gravitational potential energy, which is given by
[tex]E_i = U = mgh[/tex]
where mg is the weigth of the rock (2200 N), while h is the height at which the rock has been released (h=15 m). If we calculate it, we get
[tex]E_i = mgh=(2200 N)(15 m)=33000 J =33 kJ[/tex]
Just before hitting the ground, the rock height is zero, so its potential energy is now zero. So the total mechanical energy of the rock now is just kinetic energy:
[tex]E_f = K_f[/tex]
however, the mechanical energy of the rock must be conserved, so
[tex]E_i = E_f[/tex]
and so we have that the kinetic energy of the rock just before hitting the ground is equal to its initial potential energy:
[tex]K_f = E_i =U_i = 33 kJ = 33000 J[/tex]
