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

PR2 = PQ2 + QR2

y2 = (120)2 + x2……..(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,

y2 = x2 + (120)2

(130)2 = x2 + 14400

x2 = 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

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