A heavy stone is thrown with a speed v from a cliff of height h ....

Let the mass of the stone be $m$.

Using the principle of conservation of mechanical energy, the total energy of the stone at the point of projection is equal to its total energy just before striking the ground.

Initial energy $E_i = \dfrac{1}{2}mv^2 + mgh$

Final energy $E_f = \dfrac{1}{2}mV^2$

Equating the initial and final energies:

$\dfrac{1}{2}mV^2 = \dfrac{1}{2}mv^2 + mgh$

$V = \sqrt{v^2 + 2gh}$

The final speed $V$ depends only on the initial speed $v$, the height $h$, and acceleration due to gravity $g$. It does not depend on the angle of projection.

Therefore, the speed with which the stone strikes the ground is independent of the initial direction.