Question

Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of water and fly at the end of the strings rests on the water 36 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string is taut, how much string does she have out?

If she pulls the string at the rate of 5 m/s, what will be the horizontal distance of the fly from her after 12 seconds?
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Answer 1

ANSWER:-

Given:-

In \triangle ABC

${AC {}^{2} = BC {}^{2} + AB {}^{2} = (2.4) {}^{2} + (1.8) {}^{2} }$

$= 5.76 = 3.24$

$\implies AC = \sqrt{9.00} = 3 \: m$

Let the length of the string she pulled out in 12 s

be BD.

Then, the new length of fly = 0.05 x 12 = 0.60 m Remaining length of fly (AD) = 3-0.60 = 2.4 m Now in ABD,

$AD{}^{2} = AB {}^{2} + BD {}^{2}$

${ \implies BD {}^{2} = AD {}^{2} - AB {}^{2} = (2.4) {}^{2} - (1.8) {}^{2} }$

$\implies BD = \sqrt{5.76 - 3.24} = 2.52$

$= 1.587 \: m \: = 1.59 \: m$

Horizontal length of the fry = (1.59 + 1.2) m

= 2.79

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Answer 2

$\underline{\underline{\large\bf{Assumption-}}}$

$\textsf{Let PR be the height of the tip of the fishing rod}$

$\textsf{from the water surface. Let QR be the horizontal}$

$\textsf{distance of the fly from the tip of the fishing rod.}$

$\textsf{Then, PQ is the length of the string.}$

$\underline{\underline{\large\bf{To\; Find -}}}$

$\bull \textsf {PQ, the length of the string out}$

$\bull \textsf {Horizontal distance of the fly from girl after 12 seconds (SR + 1.2 m)}$

$\underline{\underline{\large\bf{Solution -}}}$

$\sf PQ^2 =PR^2 + QR^2$

$\sf PQ^2 = (1.8)^2+(2.4)^2$

$\sf PQ^2 = 3.24 + 5.76$

$\sf PQ^2 = 9.00$

$\sf PQ = 3 m$

$\boxed{\textbf\red{The length of the string out is \green{3 m}}}$

$\underline{\bf{Now,}}$

$\text{She pulls the string at the rate of 5cm/s}$

$\textsf{∴ String pulled in 12 second}$

$\sf{ = 12×5}$

$\sf{=60cm}$

$\sf{=0.6m}$

$\text{Let the fly be at point S after 12 seconds.}$

$\text{Length of string out after 12 second is PS}$

$\textsf{ PS = PQ - String pulled by Nazima in 12 seconds}$

$\textsf{ PS = 3-0.6}$

$\textsf{ PS = 2.4 m}$

$\underline{\texttt{Now in △PSR}}$

$\sf{PR^2 + SR^2 = PS^2}$

$\sf{ (1.8)^2+SR^2 = (2.4)^2}$

$\sf{3.24 + SR^2 = 5.76}$

$\sf{SR^2 = 5.76 - 3.24}$

$\sf{SR^2 = 2.52 m^2}$

$\sf{SR ≈ 1.587 m}$

$\textsf {Horizontal distance of fly =SR+1.2m}$

$\textsf{ Horizontal distance of fly =1.587m+1.2m}$

$\boxed{\textbf\purple{∴ Horizontal distance of fly =\red{2.79m}}}$