C

To find the maximum angle at which a stone can be thrown and still always move away from the thrower, we need to analyze the projectile motion of the stone.
Step 1: Understanding the conditions
When a stone is thrown at an angle \( \theta \) to the horizontal, its horizontal and vertical components of velocity can be expressed as:
\( v_x = v_0 \cos(\theta) \) (horizontal component)
\( v_y = v_0 \sin(\theta) \) (vertical component)
For the stone to always move away from the thrower, the horizontal component \( v_x \) must remain positive because motion away implies positive displacement in the horizontal direction.
Step 2: Identifying the maximum angle
The stone will reach its highest point when the vertical component of its velocity becomes zero. At this point, the stone has to be vertically above the thrower still moving horizontally away. The angle at which this condition is satisfied is 45 degrees because any angle greater than 45 degrees would result in a downward vertical component, pulling it back towards the thrower.
Step 3: Conclusion
The maximum angle to the horizontal at which the stone can be thrown and still be moving away from the thrower is therefore 45 degrees.
Therefore, the correct answer is option C.