Given: a boy of n height 1.5 m is flying a kite at a height of 151.5 m. The kite is moving with a speed of 10m/s. And the kite is 250 m away from the boy.

To find the speed at which the string is let out

Explanation: the below figure shows the above situation,

From the above figure,

Height of the kite, H = AD = 151.5 m

Height of the boy, b = BC = 1.5 m

Distance between kite and boy, x = CD = BE = 250 m

And BA is the length of the string = y

So, we need to find out the rate of increase of the string

From figure, h = AE

= AD-ED

= 151.5-1.5

= 150m

From figure it is clear that ΔABE forms right-angled triangle

Now applying the Pythagoras theorem, we get

AB² = BE²+AE²

Or y² = x²+h²………..(i)

Now substituting the corresponding values, we get

y² = (250)²+(150)²

⇒ y² = 62500+22500

⇒ y² = 85000

⇒ y = 291.5m………..(ii)

Now differentiate equation (i) with respect to time, we get

Applying the sum rule of the differentiation, we get

Now the height is not increasing so it is constant, so

Applying the derivative with respect to t, we get

Now given the kite is moving with a speed of 10m/s, so

Now substituting corresponding value in equation (iii), we get

Hence the string is let out at a rate of 8.6m/s.