The boy whose eye level is 1.3m from the ground, spots a balloon moving with the wind in a horizontal line at some height from the ground. The angle of elevation of the balloon from the eyes of the boy at an instant is 60^{0}. After 2 seconds, the angle of elevation reduces to 30^{0}. If the speed of the wind is 29√3m/s,then find the height of the balloon from the ground.

Hint : Distance covered by the balloon in 2sec = 29√3 x 2 = 58√3m

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.

**
**