heights and distances
Asked by
| 7th Feb, 2010,
04:10: PM
Let the speed of the jet plane be V in m/s and X be the initial horizontal distance of jet plane from the observer.
D, the distance it travels in 8 seconds = 8V.
tan 60 = 18003/X
or X = 18003/tan 60 = 1800 m
The horizontal distance, X increases by D = 8V in next 8 seconds and the angle of elevation becomes 45.
Hence,
tan 45 = 18003/(X+8V)
X + 8V = 18003/tan 45 = 1800
3
8V = 1800x1.73 - X = 1800x0.73 = 1314 m
V = 1314/8 = 164.25 m/s
Regards,
Team,
TopperLearning.
Answered by
| 7th Feb, 2010,
04:34: PM
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