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Question
A kite is moving horizontally at a height of 151.5 meters. If the speed of 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 boy is 1.5 m.
Solution
Given that height of the kite (h) = 151.5 m
Speed of the kite(V) = 10 m/s
Let FD be the height of the kite and AB be the height of the boy.
Let AF = xm
∴ BG = AF = xm
And `"dx"/"dt"` = 10 m/s
From the figure, we get that
GD = DF – GF
⇒ DF – AB
= (151.5 – 1.5) m = 150 m ......[∵ AB = GF]
Now in ΔBGD,
BG2 + GD2 = BD2 ......(By Pythagoras Theorem)
⇒ x2 + (150)2 = (250)2
⇒ x2 + 22500 = 62500
⇒ x2 = 62500 – 22500
⇒ x2 = 40000
⇒ x2 = 40000
Let initially the length of the string be y m
∴ In ΔBGD
BG2 + GD2 = BD2
⇒ x2 + (150)2 = y2
Differentiating both sides w.r.t., t, we get
⇒ `2x * "dx"/"dt" + 0 = 2y * "dy"/"dt"` ......`[because "dx"/"dt" = 10 "m"/"s"]`
⇒ 2 × 200 × 10 = 2 × 250 × `"dy"/"dt"`
∴ `"dy"/"dt" = (2 xx 200 xx 10)/(2 xx 250)` = 8 m/s
Hence, the rate of change of the length of the string is 8 m/s.
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