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.) ? If she pulls in the string at the rate of 5 cm per second, what will the horizontal distance of the fly from her after 12 seconds ?

Solution :

Let O be the position of nazima at the time of fishing.

Let OA be the rod and AB be the string.

Height of the tip of rod above sea level.

i.e.    AC = 1.8 m

The distance of the fly from the girl = 3.6 m

i.e.   OB = 3.6 m

In triangle ABC,

\({AB}^2\) = \({CB}^2\) + \({AC}^2\) = \((3.6 – 1.2)^2\) + \((1.5)^2\)

= 5.76 + 3.24 = 9 m

\(\implies\)   AB = 3 m

The rate of pulling the string = 5 cm/sec.

In 12 sec, the girl will pull = 12 \(\times\) 5 = 60 cm

The remaining length of the string AD = 3 – 0.6 = 2.4 m

New position of the string is AD.

In right triangle ACD,

\({CD}^2\) = \({AD}^2\) – \({AC}^2\)

= 5.76 – 3.24 = 2.52

CD = 1.6 m

OD = distance of the fly from her

\(\implies\)  OD = OC + CD = 1.2 + 1.6 = 2.8 m

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