Final answer:
The magnitude of the ball's acceleration when it leaves the gun, shot straight up, is 1g, or 9.8 m/s², in the downward direction, as that is the acceleration due to gravity acting on the ball.
Step-by-step explanation:
The magnitude of the ball's acceleration as it leaves the gun can be calculated using kinematic equations, under the assumption that it is shot straight up in a vacuum where the only force acting on it after it leaves the gun is gravity. Since it's moving upward against gravity, it will be decelerating at a rate of -g until it reaches zero velocity at its highest point.
The acceleration of any object in free fall or launched vertically upwards is at a magnitude of g, which is approximately 9.8 m/s². Therefore, the magnitude of the acceleration of the ball when it leaves the gun is 1g, or exactly 9.8 m/s², but in the opposite direction of the ball's initial velocity due to gravity pulling it back toward Earth. It's important to note that the effect of air resistance is typically ignored in these types of problems for simplification unless specifically stated otherwise.