Question

the angle of elevation of a jet plane from a point A on the ground is 60 degrees after a flight of 15 seconds,the angle of elevation changes to 30 degrees.if the jet plane is flying at a constant height of 1500 root 3 meter find the speed of a jet plane.
PLEASE VERY URGENT

Answer 1

 (Draw diagram based on gives conditions. mark E , D,C  points down and A,B points up.)

Tan 60 = AD/ED

$\sqrt3$ = 1500$\sqrt3$/ED

ED = 1500

Tan 30 = BC/EC

1/$\sqrt3$ = 1500√3/EC

DE = EC - ED

4500-1500

= 3000

PLANE TRAVELS 3000 m  in 15 sec.

speed = Dstance/Time

= 3000/15

200 m/s.

Jet plane speed = 200m/s.

Answer 2

Let height of air craft = h = 1500√3 and ground distance= d
Then we have tan 60 = h/d
                             √3 = 1500√3/d
                         ⇒  d =1500
The ground distance is  1500m
Let the  ground distance after 15 second is d1
  We have tan 30 =1500 √3/d1
              ⇒  1/√3 = 1500√3/d1
             ⇒     d1 = 4500
The ground distance is 4500m
The plane has traveled d1-d = 4500-1500=3000 m
Plane has traveled  3000 m  in 15 sec.
Hence the speed = distance/time
                           = 3000/15
                           = 200 m/s   or  200*3600/1000 =720 km/h
Speed of Jet plane is 720 km/h