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.

Answer 1

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.