Final answer:
The magnitude of the velocity of a 9 kg bowling ball dropped from a height of 25 meters just before it hits the sidewalk is approximately 22.1 m/s, calculated using the formula for energy conservation in the absence of air resistance.
Step-by-step explanation:
The problem you've described is asking about the magnitude of the velocity of a 9 kg bowling ball as it is about to hit the sidewalk after being dropped from a specific height, with air resistance neglected. This is a classic physics problem that applies the principles of kinematics and energy conservation.
To solve for the velocity of the bowling ball just before impact, we can use the formula derived from the conservation of mechanical energy, given that only gravitational potential energy is being converted to kinetic energy. The formula is v = sqrt(2gh), where g is the acceleration due to gravity, and h is the height from which the ball was dropped. Substituting g = 9.81 m/s^2 (the standard acceleration due to gravity) and h = 25 m (the height of the fall), we can calculate the velocity just before impact.
Performing the calculation: v = sqrt(2 * 9.81 m/s^2 * 25 m), we find that the velocity v is approximately 22.1 m/s.