Q. 255.0( 2 Votes )

# A kite is 120 m h

Answer :

Let P be the position of the kite and PR be the position of the string. Let QR = x and PR = y. Then from figure by applying Pythagoras theorem, we get

PR^{2} = PQ^{2} + QR^{2}

⇒ y^{2} = (120)^{2} + x^{2}……..(i)

Differentiating the above equation with respect to time, we get

Now the kite is moving away horizontally at the rate of 52m/sec, so , so the above equation becomes

…..(ii)

So when the string is 130m, y = 130m equation (i) become,

y^{2} = x^{2} + (120)^{2}

⇒ (130)^{2} = x^{2} + 14400

⇒ x^{2} = 16900 - 14400 = 2500

Therefore x = 50m

Substituting the value of x and y in equation (ii), we get

Hence the rate at which the string is being paid out is 20m/sec

Rate this question :

The side of an eqMathematics - Board Papers

For the curve, <sMathematics - Board Papers

The money to be sMathematics - Board Papers

<span lang="EN-USMathematics - Board Papers

The side of an eqMathematics - Board Papers

The volume of a sMathematics - Board Papers