A heavy stone is thrown from a cliff of height h with a speed v. The stone will hit the ground with maximum speed if it is thrown.

In all cases, the speed of the stone will be the same when it hit the ground because the speed at the top of the cliff is the same.

When all the particles collide with the ground they have the same gain in kinetic energy 

total energy at the top = total energy at ground

$$\begin{gathered} \frac{1}{2}m{u}^2+mgh=\frac{1}{2}m{v}^2 \\ {u}^2+2gh=v^2 \\ v=\sqrt{u^2+2gh} \end{gathered}$$

So the speed will not depend on the initial direction of velocity.