(Q004) During World War II, the military imaged the seafloor by sending pulses of sound waves down through the water and measuri
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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.