Two physics students are playing catch with a baseball in the Polonski Common. One throws the ball to the other but overshoot and the ball lands on the library roof instead. The ball left the student's hand 2.22 m above the ground at a speed of 17.3 m/s and an angle of 62.5o. The library is 10.7 m high and the ball lands on it while on its way down. How long was the ball in the air?

To solve this problem, we need to break it down into two parts: the time it takes for the ball to reach its maximum height, and the time it takes for the ball to fall from its maximum height to the library roof.

1. First, we need to find the vertical component of the initial velocity. We can do this using the equation V_y = V * sin(θ), where V is the initial speed and θ is the angle. Plugging in the given values, we get V_y = 17.3 m/s * sin(62.5) = 15.1 m/s.

2. Next, we need to find the time it takes for the ball to reach its maximum height. We can use the equation t = V_y / g, where g is the acceleration due to gravity (9.8 m/s^2). So, t = 15.1 m/s / 9.8 m/s^2 = 1.54 s.

3. Now, we need to find the maximum height the ball reaches. We can use the equation h = V_y * t - 0.5 * g * t^2. Plugging in the values we found, we get h = 15.1 m/s * 1.54 s - 0.5 * 9.8 m/s^2 * (1.54 s)^2 = 11.6 m. But the ball was thrown from 2.22 m above the ground, so the maximum height above the ground is 11.6 m + 2.22 m = 13.82 m.

4. Finally, we need to find the time it takes for the ball to fall from its maximum height to the library roof. We can use the equation t = sqrt((2 * h) / g), where h is the height the ball falls. The height the ball falls is 13.82 m - 10.7 m = 3.12 m. So, t = sqrt((2 * 3.12 m) / 9.8 m/s^2) = 0.8 s.

Adding the two times together, the ball was in the air for 1.54 s + 0.8 s = 2.34 s.