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Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string

Q10   Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface
          of the water and the fly at the end of the string rests on the water 3.6 m away and
          2.4 m from a point directly under the tip of  the rod. Assuming that her string
         (from the tip of her rod to the fly) is taut, how much string does she have out
          (see Fig. 6.64)? If she pulls in the string at the rate of 5 cm per second, what will be
          the horizontal distance of the fly from her after 12 seconds?

          

Answers (1)
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Let AB = 1.8 m

BC is horizontal distance between fly to the tip of the rod.

Then, the length of the string is AC.

In \triangle ABC, using Pythagoras theorem

AC^2=AB^2+BC^2

\Rightarrow AC^2=(1.8)^2+(2.4)^2

\Rightarrow AC^2=3.24+5.76

\Rightarrow AC^2=9.00

\Rightarrow AC=3 m

Hence, the length of the string which is out is 3m.

If she pulls in the string at the rate of 5cm/s, then the distance travelled by fly in 12 seconds.

                          = 12\times 5=60cm=0.6m

Let D be the position of fly after 12 seconds.

Hence, AD is the length of the string that is out after 12 seconds.

Length of string pulled in by nazim=AD=AC-12

                                                             =3-0.6=2.4 m

 

In \triangle ADB,

AB^2+BD^2=AD^2

\Rightarrow (1.8)^2+BD^2=(2.4)^2

\Rightarrow BD^2=5.76-3.24=2.52 m^2

\Rightarrow BD=1.587 m

Horizontal distance travelled by fly = BD+1.2 m

                                                       =1.587+1.2=2.787 m

                                                         = 2.79 m

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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