The angle of elevation of a jet plane from a point A on the grund is `60^0` . After and flight of 30 seconds, the angle of elevation changes to `30^0dot` If the jet plane is flying at a constant height of `3600sqrt(3)m ,` find the speed of the jet plane.

Answer

The speed of the jet plane is 864 km/h

Explanation

Suppose O is the position of the observer on the ground . A and B are the jet plane during the interval of 30 seconds .
Draw AC and BD perpendiculars to the ground.
OCD is the horizontal line on the ground
` angle AOC = 60^(@) , angle BOD = 30^(@)`
`AC = BD = PO = 3600 sqrt(3) m `
In ` Delta AOC = tan 60^(@) = (AC)/(OC)`
` :. sqrt(3)= (3600sqrt(3))/(OC)" " :. OC = 3600 ` m ...(1)
n `Delta BOD, tan angle BOD = tan 30^(@) = (BD)/(OD)`
` :. 1/sqrt(3) (3600 sqrt(3))/(OD) " " :. OD = 3600 sqrt(3) xx sqrt(3) ` m
` :. OD = ( 3600 xx 3) `m
` AB = CD = OD -OC" " ...( O -C - D)`
` :. A B = 3600 xx 3 - 3600 m " "` ...[ From (1) and (2) ]
` = 3600 (3-1) m`
` :. AB = 3600 xx 2` m
This distance AB is covered in 30 seconds .
` :. " speed of the jet plane " = (3600 xx2)/(30) m//s`
= 240 m/s
` = 240/1000 xx 60 xx 60` km//h= 864 km/h