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 2.28 m above the ground at a speed of 13.3 m/s and an angle of 63.0o. The library is 5.94 m high and the ball lands on it while on its way down. How long was the ball in the air?
Question
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 2.28 m above the ground at a speed of 13.3 m/s and an angle of 63.0o. The library is 5.94 m high and the ball lands on it while on its way down. How long was the ball in the air?
Solution
To solve this problem, we need to break it down into two parts: the vertical motion and the horizontal motion.
- Vertical motion:
First, we need to find the time it takes for the ball to reach its maximum height. We can use the equation of motion: v = u + at, where v is the final velocity (0 m/s at the maximum height), u is the initial vertical velocity, a is the acceleration (gravity, which is -9.8 m/s²), and t is the time.
The initial vertical velocity can be found using the equation u = u0 * sin(θ), where u0 is the initial speed (13.3 m/s) and θ is the angle (63.0°).
u = 13.3 m/s * sin(63.0°) = 11.6 m/s
Now we can find the time it takes to reach the maximum height:
0 = 11.6 m/s - 9.8 m/s² * t
t = 11.6 m/s / 9.8 m/s² = 1.18 s
- The total time in the air is the time it takes to reach the maximum height plus the time it takes to fall from the maximum height to the library roof.
The height difference between the maximum height and the library roof can be found using the equation h = u * t + 0.5 * a * t², where h is the height, u is the initial vertical velocity (0 m/s when falling from the maximum height), a is the acceleration (-9.8 m/s²), and t is the time.
The height difference is the height of the library roof minus the height at which the ball was thrown: 5.94 m - 2.28 m = 3.66 m
Now we can find the time it takes to fall this distance:
3.66 m = 0.5 * -9.8 m/s² * t²
t² = 3.66 m / 0.5 * -9.8 m/s² = 0.75 s²
t = sqrt(0.75 s²) = 0.87 s
- The total time the ball was in the air is the time to reach the maximum height plus the time to fall to the library roof:
1.18 s + 0.87 s = 2.05 s
So, the ball was in the air for approximately 2.05 seconds.
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