Question

A boy , whose eye level is 1.3m from the ground , spots a ballon moving with the wind in a horizontal line at same height from the ground. the angle of elevation of the ballon from the eyes of the boy at any instant is 600 . after 2 seconds , the angle of elevation reduces to 300 if the speed of the wind at that moment is 29√ m/s , then find the height of the ballon from the ground .

Answer 1

let the boy spot the balloon at point A.

and the angle of elevetion from his eye point P to the balloon is 60 deg.

PQ=1.3 m.

let after 2 seconds the balloon reaches to the point B. and now its angle of elevation from the eye is 30 deg.

QCF is the horizontal line at the ground level.

since speed of the wind =

the distance covered by the balloon in 2 sec be AB=.........(a)

in the right angled triangle APD,

.............(1)

in the right angled triangle BEP,



PE=PD+DE

PE=PD+AB [since DE=AB]

[using (a)]

from (1) and (2), substitute the values in terms of AD.



therefore the height of the balloon from the ground level is AC=AD+DC

AC=AD+PQ

AC=87+1.3=88.3 m.

thus the height of the balloon from the ground is 88.3 m.

Answer 2

Answer:

above is correct but plz

Explanation:

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