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
In this way, the pace of progress of the length of the string is \[8\text{ }m/s.\]