A kite is moving horizontally at a height of `151.5 m`. If the speed of the kite is `10 m/s`, how fast is the string being let out, when the kite is 250 m away from the boy who is flying the kite? The height of the boy is 1.5 m. (A) `8` m/s (B) `12 ` m/s (C) `16` m/s (D) `19` m/s

Explanation

We have, height(h)=151.5m, speed of kite (V) = 10m/s
Let CD be the height of kite and AB be the height of boy.
Let DB=xm= EA and AC = 250m
`therefore (dx)/(dt)=10m/s`
From the figure, we see that
EC=151.5-1.5=150m
and AE=x
Also, AC=250m

In right angled `DeltaCEA`,
`AE^(2) + EC^(2)= AC^(2)`
`rArr x^(2) +(150)^(2)=y^(2)`
`rArr x^(2) + (150)^(2)=(250)^(2)`
`rArr x^(2) = (250)^(2) - (150)^(2)`
`=(250+150)(250-150)`
`=400 xx 100`
`therefore x=20 xx 10=200`
From Eq.(i), on differentiating w.r.t.t, we get
`2x.(dx)/(dt) +0 = 2y(dy)/(dt)`
`rArr 2y(dy)/(dt) = 2x(dx)/(dt)`
`therefore (dy)/(dt) = x/y.(dx)/(dt)`
`=200/250.10=8 m//s` `[therefore (dx)/(dt) = 10m//s]`
So, the required rate at which the string is being let out is 8 m/s.