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Answer :
Final answer:
According to the calculations, the water balloon does not hit the professor. It comes closest to her when it is approximately 0.316 meters away.
Explanation:
In order to determine if the water balloon hits the professor, we need to calculate the time it takes for the balloon to reach the ground. The time can be found using the equation:
t = \sqrt{\frac{2h}{g}}
Where t is the time, h is the height (18.0 meters), and g is the acceleration due to gravity (9.8 m/s^2). Plugging in the values, we find that it takes approximately 1.52 seconds for the balloon to reach the ground.
We also need to calculate the horizontal distance the professor walks in that time. The distance can be found using the equation:
d = vt
Where d is the distance, v is the velocity (0.450 m/s), and t is the time (1.52 seconds). Plugging in the values, we find that the professor walks approximately 0.684 meters during that time.
Therefore, since the professor is 1.00 meter from the point directly beneath the window, the balloon does not hit her. It comes closest to her when it is 0.316 meters away from her.
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Rewritten by : Jeany
Refer to the diagram shown below.
In order for the balloon to strike the professor's head, th balloon should drop by 18 - 1.7 = 16.3 m in the time at the professor takes to walk 1 m.
The time for the professor to walk 1 m is
t = (1 m)/(0.45 m/s) = 2.2222 s
The initial vertical velocity of the balloon is zero.
The vertical drop of the balloon in 2.2222 s is
h = (1/2)*(9.8 m/s²)*(2.2222 s)² = 24.197 m
Because 24.97 > 16.3, the balloon lands in front of the professor, and does not hit the professor.
The time for the balloon to hit the ground is
(1/2)*(9.8)*t² = 18
t = 1.9166 s
The time difference is 2.2222 - 1.9166 = 0.3056 s
Within this time interval, the professor travels 0.45*0.3056 = 0.175 m
Therefore the balloon falls 0.175 m in front of the professor.
Answer:
The balloon misses the professor, and falls 0.175 m in front of the professor.
In order for the balloon to strike the professor's head, th balloon should drop by 18 - 1.7 = 16.3 m in the time at the professor takes to walk 1 m.
The time for the professor to walk 1 m is
t = (1 m)/(0.45 m/s) = 2.2222 s
The initial vertical velocity of the balloon is zero.
The vertical drop of the balloon in 2.2222 s is
h = (1/2)*(9.8 m/s²)*(2.2222 s)² = 24.197 m
Because 24.97 > 16.3, the balloon lands in front of the professor, and does not hit the professor.
The time for the balloon to hit the ground is
(1/2)*(9.8)*t² = 18
t = 1.9166 s
The time difference is 2.2222 - 1.9166 = 0.3056 s
Within this time interval, the professor travels 0.45*0.3056 = 0.175 m
Therefore the balloon falls 0.175 m in front of the professor.
Answer:
The balloon misses the professor, and falls 0.175 m in front of the professor.
