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Answer this question or recommend an answer
Physics Question. Some would be appreciated.
Experiencing a constant horizontal 1.10 m/s wind, a hot-air balloon ascends from the launch site at a constant vertical speed of 2.50 m/s. At a height of 205 m, the balloonist maintains constant altitude for 10.0 s before releasing a small sandbag. How far from the launch site does the sandbag land?
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- By livewoutregret16 |
- Category: physics |
- Grade level: 12th grade (18)
- Nov/12/2008 |
- Answers to this question: 1
|
Answer by: Ganesha
First Answer
Time to reach 205 m = 205/2.5 =82s
The horizontal distance in 82 s = 82*1.1 = 90.2.
The horizontal distance in anothe 10 s of maitained altitude of 205 = 10*1.1 = 11.
The total horizontal distance = 90.2+11 =101.2m-------------(1)
The the time t to reach the ground is given by H =0.5gt^2==>
205 =0.5*gt^2
t =sqrt(205*2/g) = 6.4648s
The horintal distance covered during ttrajectory movement of the bag = horizontal velocity+ time to reach the ground
=1.1*6.4648 =7.1113---------------------------------------------------(2)
Therefore, the horizotal distance of landing from the place oflaunch = (1)+(2) = 101.2+7.113 =108.3113m
The horizontal distance in 82 s = 82*1.1 = 90.2.
The horizontal distance in anothe 10 s of maitained altitude of 205 = 10*1.1 = 11.
The total horizontal distance = 90.2+11 =101.2m-------------(1)
The the time t to reach the ground is given by H =0.5gt^2==>
205 =0.5*gt^2
t =sqrt(205*2/g) = 6.4648s
The horintal distance covered during ttrajectory movement of the bag = horizontal velocity+ time to reach the ground
=1.1*6.4648 =7.1113---------------------------------------------------(2)
Therefore, the horizotal distance of landing from the place oflaunch = (1)+(2) = 101.2+7.113 =108.3113m
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- Nov/13/2008 |
- Answers by Ganesha: 1193 |
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