Final answer:
The time of flight for a baseball thrown from 1st base to 2nd base, reaching a maximum height of 2.77 m, is approximately 1.504 seconds.
Step-by-step explanation:
To find the time of flight of a baseball thrown from 1st base to 2nd base, where it reaches a maximum height of 2.77 m and lands at the same height from which it was thrown, we can use the equations of motion for projectile motion.
Since the height the ball starts and ends at the same level, we can use the formula for the time to reach maximum height and then double it, because time of ascent equals time of descent for symmetrical trajectories.
The equation to calculate the time (t) to reach the maximum height (h) is derived from h = (1/2)g*t^2, where g is the acceleration due to gravity (approximately 9.81 m/s^2). Rearranging the equation to solve for t gives us t = √(2h/g). Plugging in the maximum height given, 2.77 m, we find t = √(2*2.77 m / 9.81 m/s^2) which gives us the time to reach the maximum height. To find the total time of flight, we simply double this value.
t = 2 * √(2*2.77 m / 9.81 m/s^2)
Upon solving, we find that the time of flight of the baseball is approximately:
t = 2 * √(2*2.77 / 9.81) ≈ 2 * 0.752 s ≈ 1.504 s
So, the total time of flight for the baseball is approximately 1.504 seconds.