A boy whose eye level is `1.3 m` from the ground, spots a balloon moving with the wind in a horizontal line at same height from the ground. The angle of elevation of the balloon from the eyes of the boy at an instant is `60^(@)`. After 2 seconds, the angle of elevation reduces to `30^(@)`. If the speed of the wind is `29sqrt(3) m//s` then find the height of the balloon from the ground.

Explanation

Let AB be the position of the boy and AX be the the horizontal grund. Let C and D be the two positions of the ballon. Draw `CL_|_AX,DM_|_AX` and `BN_|_DM`, intersecting CL at P. Then `angleCBN= 60^(@)` and `angleDBN = 30^(@)`.
Distance covered by the balloon in 2 seconds.
`=(29sqrt(3) xx 2) m = 58sqrt(3) m`.
`:' CD = 58sqrt(3)`.
Let `BP = x` metres, Then,
`BN = (BP + PN) = (BP + CD)= (x+ 58sqrt(3)) m and DN = CP` From right `Delta BND`, we have
`(DN)/(BN) = tan 30^(@) = 1/(sqrt(3))`.
`rArr (DN)/((x+58sqrt(3))m) = 1/(sqrt(3)) rArr DN = ((x+58sqrt(3)))/(sqrt(3)) m"........"(i)`
From right `DeltaBPC`, we have
`(CP)/(BP) = tan 60^(@) = sqrt(3)`
`rArr(CP)/(x m) = sqrt(3) rArr CP = (xsqrt(3))`
`rArr (3x-x) = 58sqrt(3) rArr 2x = 58 sqrt(3) rArr x = 29 sqrt(3)`.
From (ii), we get
`CP =(29sqrt(3) xx sqrt(3)) m = (29 xx 3) m = 87 m`
`:.` height of the ballon from the ground.
`= CL = CP + PL= CP + AB = 87 m + 1.3 m = 88.3 m`
Hence, the height of the balloon from the ground is `88.3` metres.