When and How Do They Land?

Both situations have an initial speed, $u$ , of 0.
But let's say that the height is 10 meters

You can use the "suvat" equations to solve for both.

For the vertical ball:
$\begin{aligned} v^2 &=& u^2 + 2as \\ v &=& = \sqrt{0^2 + 2as} \\ &=& \sqrt{2 \cdot 9.81 \cdot 10} \\ &\approx& 14.00 \text{ m/s} \end{aligned}$

And to find the time taken:

$\begin{aligned} v &=& u + at \\ &=& 0 + at \\ t &=& \frac va \\ &=& 14/9.81 \\ &\approx& 1.43 \text{ s} \end{aligned}$

For the horizontal ball,
there are 2 components therefore it is a different speed
$v^2_{x}=u^2_{x} + 2as_{x}$ horizontal component.
$v^2_{y}=u^2_{y} + 2as_{y}$ vertical component.

In both cases, initial vertical component of velocity is zero. So both reaches the ground simultaneously but with different speeds as in second case there is horizontal component of velocity.