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Home » 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.
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.
Application of Derivatives | Class 12 | Exercise 1 | Maths | NCERT Exemplar | NCERT Exemplar
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.

NCERT Exemplar Solutions Class 12 Mathematics Chapter 6 - 3

Speed of the kite(V)

    \[=\text{ }10\text{ }m/s\]

Leave FD alone the tallness of the kite and AB be the stature of the kite and AB be the tallness of the kid.

Presently, let AF

    \[=\text{ }x\text{ }m\]

Along these lines,

    \[BG\text{ }=\text{ }AF\text{ }=\text{ }x\]

What’s more,

    \[dx/dt\text{ }=\text{ }10\text{ }m/s\]

From the figure, it’s seen that

    \[GD\text{ }=\text{ }DF\text{ }\text{ }GF\text{ }=\text{ }DF\text{ }\text{ }AB\]

    \[=\text{ }\left( 151.5\text{ }\text{ }1.5 \right)\text{ }m\text{ }=\text{ }150\text{ }m\text{ }\left[ As\text{ }AB\text{ }=\text{ }GF \right]\]

Presently, in ∆ BDG

    \[BG2\text{ }+\text{ }GD2\text{ }=\text{ }BD2\text{ }\left( By\text{ }Pythagoras\text{ }Theorem \right)\]

    \[x2\text{ }+\text{ }\left( 150 \right)2\text{ }=\text{ }\left( 250 \right)2\]

    \[x2\text{ }+\text{ }22500\text{ }=\text{ }62500\]

    \[x2\text{ }=\text{ }62500\text{ }\text{ }22500\text{ }=\text{ }40000\]

    \[x\text{ }=\text{ }200\text{ }m\]

Leave at first the length of the string alone y m

Along these lines, in ∆ BDG

    \[BG2\text{ }+\text{ }GD2\text{ }=\text{ }BD2\]

    \[x2\text{ }+\text{ }\left( 150 \right)2\text{ }=\text{ }y2\]

Separating the two sides w.r.t., t, we have

NCERT Exemplar Solutions Class 12 Mathematics Chapter 6 - 4

In this way, the pace of progress of the length of the string is

    \[8\text{ }m/s.\]

i

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