Q. 145.0( 3 Votes )


An aeroplane flying horizontally at height of 2500 3 mtsabove that ground; is observed to be at angle of elevation 60° from theground. After a flight of 25 seconds the angle of elevation is 30° . Findthe speed of the plane in m/sec.

Answer :


https://gs-post-images.grdp.co/user_files/63094/15047/images/extra-9_files/gtd.gif

Given:

Angle of elevation, θ = 30°

Height above the ground = 2500√ 3 m

To find: speed of the plane

Solution:


Let P and Q be the two positions of the aeroplane.


QB = 2500
√ 3 m


Let the speed of the aeroplane be s m/sec.


tan 60° = 2500√ 3 / x


(2500√ 3 / x) = √ 3

x = 2500 m


tan 30° = (
2500√ 3) / ( x + y)

( 2500√ 3) / ( x + y) = 1 √ 3

x + y = 2500 √ 3 × √ 3


= 2500 × 3


= 7500

y = 7500 – 2500


= 5000 m


To travel 5000 m the aeroplane takes 25 seconds.

The speed of the plane = 5000 / 25

= 200 m/sec.

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