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An aircraft is flying at a height of 340...

An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is 30°, what is the speed of the aircraft ? (tan = 15° = 0.2679)

Text Solution

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Let O be the ground observation point and A, B, C be the positions of the aircraft at t = 0s, t = 5s and t = 10s respectively,

Clearly, `tan15^(@) =(AB)/(OB)` or `AB =OB tan 15^(@)`
Thus, distance travelled by the aircraft in 10 s.
i.e., `s=AC =2AB =2(OB tan 15^(@))`
`=2 xx 3400 xx 0.2679 = 1822`m
Speed of the aircraft `=s/t =(1822 m)/(10 s) = 182.2` m/s
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