Final answer:
Using projectile motion principles, the bullet's flight time was calculated to be roughly 0.0207 seconds and the speed at which it left the barrel was calculated to be 2270 m/s. However, these numbers appear to be inconsistent with real-world scenarios.
Step-by-step explanation:
To calculate the bullet's flight time and the speed it left the barrel, we can use the principles of projectile motion. Given that the bullet hits 2.10 cm below the aim point and was aimed horizontally, the only force acting on the bullet in the vertical direction is gravity. The horizontal distance (range) does not affect the time it takes for the bullet to fall this vertical distance.
Using the formula for the vertical motion Y = 0.5 * g * t^2, where Y is the vertical distance the bullet fell (2.10 cm or 0.021 m), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time in seconds, we can solve for t:
0.021 = 0.5 * 9.8 * t^2
t≈0.0207 seconds
To find the speed it left the barrel, we utilize the horizontal motion formula: speed = distance/time. We have the horizontal distance (47.0 m) and now the time (0.0207 s), so:
speed = 47.0 m / 0.0207 s
speed≈2270 m/s
Please note that the numbers are not adding up logically, as a bullet traveling for only about 0.0207 seconds cannot cover 47.0 meters, indicating that either the data provided or the calculations may be inaccurate or not based on real-world scenarios.