Question

The height of a hill is 270m above the level of a horizontal plane. From a point A on this plane the angular elevation of the top of the hill is 600 . A balloon rises from A and ascends vertically upwards at a uniform rate; after 2.5 minutes the angular elevation of the top of the hill to an observer in the balloon is 300 . Find the speed of the balloon’s ascent in m/s.

Answer 1

Answer:

Step-by-step explanation:

Let CD be the hill.∴CD=270 m⇒∠ACD=60°In △ACD,tan∠ACD=CDAC⇒tan60°=270AC⇒3–√=270AC⇒AC=2703√=903–√ m⇒BE=903–√ mIn △DEB,tan∠EBD=DEBE⇒tan30°=DE903√⇒13√=DE903√⇒DE=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