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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