A

Step 1: Analyze the projectile motion when thrown at an angle of 30° to the horizontal.
The horizontal range (R) is given by the formula:
$$ R = \frac{v^2 \sin(2\theta)}{g} $$
where \( v \) is the initial speed, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity (approximately 9.81 m/s²).
For the first case, \( \theta = 30° \):
$$ R = \frac{v^2 \sin(60°)}{g} = \frac{v^2 \cdot \frac{\sqrt{3}}{2}}{g} $$
Given that R = 90 m, we can derive an expression relating v and g.

Step 2: Analyze the second case when the ball is thrown at an angle of 30° to the vertical. This means the launch angle with respect to the horizontal becomes 60°.
Therefore, we can use the range formula again:
$$ R' = \frac{v^2 \sin(120°)}{g} = \frac{v^2 \cdot \frac{\sqrt{3}}{2}}{g} $$ = \frac{v^2 \cdot \frac{\sqrt{3}}{2}}{g} \text{, which is equivalent to the expression for } R \text{ in the first case but with } \sin(120°) \text{ instead of } \sin(60°).\
The sine of 120° is the same value but in a different context: \( \sin(120°) = \sin(180° - 60°) = \sin(60°) \).
Step 3: Since both angles result in the same form of the range formula and given the initial speed was the same, we can compare the ranges: The first range at 30° is already known to be 90 m. The angle of projection leading to a greater time of flight increases the range proportionally considering symmetrical projectile motion is maintained.
Therefore, the range at 60° becomes:
R' = 90 * 2 = 120 m.

Therefore, the answer is 120 m.