A kite is moving horizontally at a height of $151.5$ meters. If the speed of kite is
10 m/s, how fast is the string being let out; when the kite is 250 m away from
the boy who is flying the kite? The height of boy is 1.5 m.
Speed of the kite(V) = 10 m/s
Let FD be the height of the kite and AB be the height of the kite and AB be the height of the boy.
Now, let AF = x m
So, BG = AF = x
And, dx/dt = 10 m/s
From the figure, it’s seen that
GD = DF – GF = DF – AB
= (151.5 – 1.5) m = 150 m [As AB = GF]
Now, in ∆ BDG
BG2 + GD2 = BD2 (By Pythagoras Theorem)
x2 + (150)2 = (250)2
x2 + 22500 = 62500
x2 = 62500 – 22500 = 40000
x = 200 m
Let initially the length of the string be y m
So, in ∆ BDG
BG2 + GD2 = BD2
x2 + (150)2 = y2
Differentiating both sides w.r.t., t, we have
$2 x \cdot \frac{d x}{d t}+0=2 y \cdot \frac{d y}{d t} \quad\left[\because \frac{d x}{d t}=10 \mathrm{~m} / \mathrm{s}\right]$
$2 \times 200 \times 10=2 \times 250 \times \frac{d y}{d t}$
$\frac{d y}{d t}=\frac{2 \times 200 \times 10}{2 \times 250}=8 \mathrm{~m} / \mathrm{s}$
Therefore, the rate of change of the length of the string is 8 m/s.