Final answer:
A baseball thrown vertically upwards at a velocity of 12 m/s is in the air for approximately 2.44 seconds, which includes the time it takes to reach the peak of its motion and to fall back to the initial height.
Step-by-step explanation:
To determine how long a baseball thrown vertically upwards at a velocity of 12 m/s is in the air, we can use the kinematic equations for uniformly accelerated motion. Since the acceleration due to gravity is 9.81 m/s2 downward, and we are ignoring air resistance, the time it takes for the ball to reach the peak of its motion is found using the equation
v = u + at,
where v is the final velocity (0 m/s at the peak), u is the initial velocity (12 m/s), a is the acceleration (-9.81 m/s2, since it is acting opposite to the initial throw), and t is the time.
Thus, the time to reach the peak is given by
0 = 12 m/s - (9.81 m/s2 * t).
Solving for t gives t = 12 m/s / 9.81 m/s2 = approximately 1.22 seconds.
But this is just the time to reach the peak, and the baseball will take the same amount of time to come back down, so we double this time to get the total time in the air:
2 * 1.22 s = approximately 2.44 seconds.