A

Step 1: Understand the problem setup.
A ball is thrown horizontally from the top of a tower 19.6 m high. It falls under the influence of gravity and we need to find the initial horizontal velocity when it hits the ground at a 45º angle with the horizontal.

Step 2: Analyze the vertical motion.
The ball is subject to gravitational acceleration, and it falls freely from a height of 19.6 m. The equation of motion for the vertical displacement $y$ is given by:
$$ y = ut + \frac{1}{2}gt^2 $$
Here, the initial vertical velocity $u = 0$, $g = 9.8 \, \text{m/s}^2$, and $y = 19.6 \, \text{m}$. Thus:
$$ 19.6 = 0 + \frac{1}{2} \cdot 9.8 \cdot t^2 $$
This simplifies to:
$$ 19.6 = 4.9t^2 $$
Dividing both sides by 4.9 gives:
$$ t^2 = 4 \Rightarrow t = 2 \, \text{s} $$

Step 3: Analyze the horizontal motion.
The horizontal distance traveled by the ball can be represented as:
$$ x = vt $$
where $v$ is the initial horizontal velocity. Therefore:
$$ x = v \cdot 2 $$

Step 4: Determine the angle of projection.
Since the ball hits the ground at a 45º angle, the horizontal and vertical distances covered by the ball are equal at the point of impact. Thus, we can say:
$$ x = y $$
From our earlier calculation, $y = 19.6 \, \text{m}$, so:
$$ x = 19.6 \, \text{m} $$
Substituting for $x$ we have:
$$ 19.6 = v \cdot 2 \Rightarrow v = \frac{19.6}{2} = 9.8 \, \text{m/s} $$

Step 5: Conclusion.
Therefore, the initial velocity of the ball is 9.8 ms^–1.

Thus, the correct answer is: A.