Question

22. 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
2.4 m -
1.2 m
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?
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A​

Answer 1

Answer:

this question was longest and I cannot understand

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.)