A kite is 120 m high and 130 m of string is | vedclass

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(B) Let $P$ be the position of the kite and $PR$ be the string. Let $PQ = 120 \ m$ be the constant height.Let $QR = x$ and $PR = y$.By the Pythagoras

Vedclass · Accepted Answer

$15 \ m/sec$.

Vedclass · Answer

(B) Let $P$ be the position of the kite and $PR$ be the string. Let $PQ = 120 \ m$ be the constant height.Let $QR = x$ and $PR = y$.By the Pythagoras theorem in $	riangle PQR$:$y^2 = (120)^2 + x^2 \dots (i)$Differentiating with respect to time $t$,we get:$2y \frac{dy}{dt} = 2x \frac{dx}{dt}$$y \frac{dy}{dt} = x \frac{dx}{dt} \dots (ii)$Given that the kite is moving away horizontally at the rate of $\frac{dx}{dt} = 39 \ m/sec$.From $(i)$,when $y = 130 \ m$:$(130)^2 = (120)^2 + x^2$$x^2 = 16900 - 14400 = 2500$$x = 50 \ m$Substituting the values into $(ii)$:$130 \frac{dy}{dt} = 50 	imes 39$$\frac{dy}{dt} = \frac{50 	imes 39}{130} = \frac{1950}{130} = 15 \ m/sec$.Thus,the rate at which the string is being paid out is $15 \ m/sec$.

$A$ kite is $120 \ m$ high and $130 \ m$ of string is out. If the kite is moving away horizontally at the rate of $39 \ m/sec$,then the rate at which the string is being paid out is:

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