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  • 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.

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.

Answers (1)

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 letting out speed of the string

Explanation: the above situation is explained by the figure,

Referring to the above figure,

Height of the kite, H = AD = 151.5 m

Height of the boy, b = BC = 1.5 m

 x = CD = BE

Distance between kite and boy = AB = y =250

So, we need to calculate the increasing rate of the string

From figure, h = AE

= AD-ED

= 151.5-1.5

= 150m

The figure implies that the ΔABE is a right-angled triangle

Applying of the Pythagoras theorem results into,

\\AB\textsuperscript{2} = BE\textsuperscript{2}+AE\textsuperscript{2}\\ $ {Or y\textsuperscript{2} = x\textsuperscript{2}+h\textsuperscript{2}$ \ldots $ $ \ldots $ $ \ldots $ ..(i)}\\ {Substitute the corresponding values}\\ {y\textsuperscript{2} = (x)\textsuperscript{2}+(150)\textsuperscript{2}}

Let’s differentiate the equation (i) with respect to time,

\frac{\mathrm{d}\left(\mathrm{y}^{2}\right)}{\mathrm{dt}}=\frac{\mathrm{d}\left(\mathrm{x}^{2}+\mathrm{h}^{2}\right)}{\mathrm{dt}}$
After using the differentiation sum rule, we get

\Rightarrow \frac{\mathrm{d}\left(\mathrm{y}^{2}\right)}{\mathrm{dt}}=\frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dt}}+\frac{\mathrm{d}\left(\mathrm{h}^{2}\right)}{\mathrm{dt}}$
since the height is not increasing, it indicates that it is constant, thus
\Rightarrow \frac{\mathrm{d}\left(\mathrm{y}^{2}\right)}{\mathrm{dt}}=\frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dt}}+0$
Let's apply the derivative with respect to t

\Rightarrow 2 \mathrm{y} \cdot \frac{\mathrm{d}(\mathrm{y})}{\mathrm{dt}}=2 \mathrm{x} \cdot \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dt}}$

\Rightarrow \mathrm{y} \cdot \frac{\mathrm{d}(\mathrm{y})}{\mathrm{dt}}=\mathrm{x} \cdot \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dt}} \ldots$(iii)
since the speed of the kite is 10 m/s so

\frac{d x}{d t}=10 \mathrm{~m} / \mathrm{s}$

When y = 250

{250\textsuperscript{2} = (x)\textsuperscript{2}+(150)\textsuperscript{2}}

x=200
After substituting the corresponding values in equation (iii), we get 

\Rightarrow(250) \cdot \frac{\mathrm{d}(\mathrm{y})}{\mathrm{dt}}=(200).(10)
\frac{\mathrm{d}(\mathrm{y})}{\mathrm{dt}}=\frac{(200).(10)}{250}=8
Therefore, the letting out speed of the string is 8 m/s

Posted by

infoexpert22

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