Question

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?​

Answer 1

Answer:

Let AB be the height of the tip of the fishing rod from the water surface.

Let BC be the horizontal distance of the fly from the tip of the fishing rod.

Then, AC is the length of the string.

AC² = AB² + BC² [ By Pythagoras theorem ]

AC² = (1.8)² + (2.4)²

AC² = 3.24 + 5.76

AC² = 9

AC=3m

Thus, the length of the string out is 3\,m3m

She pulls the string at the rate of 5cm/s

∴ String pulled in 12 second = 12×5

= 60cm

= 0.6m

Let the fly be at point D after 12 seconds.

Length of string out after 12 second is AD.

AD = AC - String pulled by Nazima in 12 seconds.

AD = (3-0.6)m

= 2.4m

In triangle ADB,

AB² + BD² = AD²

(1.8)² + BD² = (2.4)²

BD² = 5.76 - 3.24

BD² = 2.52

∴ BD = 1.587m

Horizontal distance of fly = BD+1.2m

= 1.587 + 1.2

= 2.79m

Answer 2

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: QUESTION \: \: } \mid}}}}}$

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

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: SOLUTION \: \: } \mid}}}}}$

$\mathfrak{Let,} \\ \: \: \: \: \: \: \: \: \: \text{ \red{AB} be the height of the rod tip from the} \\ \text{surface of water and \red{BC} be the horizontal } \\ \text{distance between fly to tip of the rod. Then, } \\ \text{\red{AC} will be the length of the string.}$

$\: \: \: \: \: \: \: \underline{ \text{ \: In Figure. 1, \: }} \\ \\ \underline{\bold{ \: \: By \: \: using \: \: Pythagoras \: \: theorem, \: \: in \: \:}} \\ \underline{ \bold{ \: the \: \red{ \triangle \: ABC}, \: }} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \bold{AC {}^{2} = AB {}^{2} + BC {}^{2}} \\ \\ \bold{\implies AC {}^{2} = (1.8 \: m) {}^{2} + (2.4 \: m) {}^{2} } \\ \\ \bold{ \implies AC {}^{2} = (3.24 + 5.76) \: m {}^{2} } \\ \\ \bold{\implies AC {}^{2} = 9 \: m {}^{2} } \\ \\ \bold{\implies AC = \sqrt{ 9 \: m {}^{2} } } \\ \\ \: \: \: \: \: \: \bold{ \therefore \: \red{AC = 3 \: m }} \\ \\ \bold{ \therefore \: The \: \: length \: \: of \: \: the \: \: string, \red{AC \: = 3 \: m}.}$

$\: \: \: \: \: \: \tt{ If \: \: Nazima \: \: pulls \: \: in \: \: the \: \: string \: \: at \: \: the \: \: } \\ \tt{rate \: \: of \: \: 5 cm/s, \: then \: \: the \: \: distance \: \: } \\ \tt{ travelled \: \: by \: \: fly \: \: in \: \: 12 \: \: seconds \: \: will \: \: be } \\ \\ \tt{= \red{( 5 \times 12) \: cm }= \red{60 \: cm} = \red{0.6 \: m}}$

$\mathfrak{Let,} \\ \: \: \: \: \: \: \: \: \text{ \red{D} be the position of fly after \red{12 seconds}. } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \text{[Figure. 2]} \\ \text{ Hence, } \\ \: \: \: \text{ \red{AD} will be the length of string which is out} \\ \text{ after \red{12 seconds}.} \\ \\ \tt{ \therefore \: Length \: of \: \: the \: \: string \: \: pulls \: \: by \: \: Nazima} \\ \\ \: \: \: \: \: \: \: \: \: \tt{ = \pink{AD} = \red{3 \: m - 0.6 \: m} = \red{2.4 \: m}}$

$\mathfrak{Now,} \\ \: \: \: \: \: \: \: \: \: \: \underline{ \text{ \: In \: Figure. 2, \: }} \\ \\ \underline{ \bold{ \: By \: \: using \: \: Pythagoras \: \: theorem \: \: in \: \: }} \\ \underline{ \bold{ \: the \: \triangle ADB, \: }} \\ \\ \: \: \: \: \: \: \: \: \: \: \bold{AD {}^{2} = AB {}^{2} + BD {}^{2} } \\ \\ \bold{ \implies (2.4 \: m) {}^{2} =(1.8 \: m) {}^{2} + BD {}^{2} } \\ \\ \bold{\implies 5.76 \: m {}^{2} = 3.24 \: m {}^{2} + BD {}^{2} } \\ \\ \bold{\implies BD {}^{2} = (5.76 - 3.24) \: m {}^{2} } \\ \\ \bold{\implies BD {}^{2} = 2.52 \: m {}^{2}} \\ \\ \bold{ \implies \: BD = \sqrt{2.52 \: m {}^{2} }} \\ \\ \: \: \: \: \: \bold{\therefore \: \pink{BD = 1.59 \: m }\: \: \: (Approx.)}$

$\sf{Hence, \: the \: \: horizontal \: \: distance \: \: of \: \: the \: \: fly \: \: } \\ \sf{ from \: \: Nazima \: \: after \: \: 12 \: seconds \:= (1.59 + 1.2 ) \: m } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{= \pink{2.79 \: m} \: (Approx.)}$