A heavy stone is projected from a cliff of height h in a specified...

Let the mass of the heavy stone be $m$, the speed of projection be $u$, and the height of the cliff be $h$.

By the principle of conservation of mechanical energy, the total energy of the stone at the point of projection is equal to the total energy just before it strikes the ground.

Initial mechanical energy, $E_i = \dfrac{1}{2}mu^2 + mgh$

Final mechanical energy, $E_f = \dfrac{1}{2}mv^2$

Equating the initial and final energies:

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

$v^2 = u^2 + 2gh$

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

From the above expression, it is clear that the final speed $v$ depends on the initial speed of projection $u$.

Since $h > 0$ and $g > 0$, we have $v^2 > u^2$, which implies $v > u$.

Thus, the speed at which the stone strikes the ground must be larger than the speed of projection.