the height of a hill is 270 m above the level of a horizontal plane . From a point A on this plane the angular elevation of the top of the hill is 60 degrees .A Balloon rises from A and ascends vertically upwards at a uniform rate; after 2.5 minutes the angularelevationof the top of the hill to an observer in the balloon is 30 degrees. Find the speed of the balloon's ascent in m/s

Let CD be the hill.CD=270 mACD=60°In ACD,tanACD=CDACtan60°=270AC3=270ACAC=2703=903 mBE=903 mIn DEB,tanEBD=DEBEtan30°=DE90313=DE903DE=90 mNow,AB=CE=CD-ED=270-90=180 mHere, baloon takes 2.5 min to reach point B from AAnd 2.5 min = 2.5×60=150 secSpeed of baloon ascent=180150=65=1.2 m/s

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