Question

8 m
M
2 Naman is doing fly-fishing in a
stream. The tip of his 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 from him and
2.4 m from the point directly under
the tip of the rod. Assuming that the
string (from the tip of his rod to the fly) is taut, how much string does
he have out (see the adjoining figure)? If he pulls in the string at the rate
of 5 cm per second, what will be the horizontal distance of the fly from
him after 12 seconds?
2.4 m
-1.2 m-​

Answer 1

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

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

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

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

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

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

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

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

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

Answer 2

Answer:

Ans. I. To find The length of AC.

By Pythagoras theorem,

AC2 = (2.4)2 + (1.8)2

AC2 = 5.76 + 3.24 = 9.00

AC = 3 m

Length of string she has out= 3 m

Length of the string pulled at the rate of 5 cm/sec in 12 seconds

= (5 x 12) cm = 60 cm = 0.60 m

Remaining string left out = 3 – 0.6 = 2.4 m

II. To find: The length of PB

PB2 = PC2 – BC2

= (2.4)2 – (1.8)2

= 5.76 – 3.24 = 2.52

PB = = 1.59 (approx.)

Hence, the horizontal distance of the fly from Nazima after 12 seconds

= 1.59 + 1.2 = 2.79 m (approx.)