Question

10. 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
the rod. Assuming that her string
2.4 m from a point directly under the tip of
(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:

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 3m

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 △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

Horizontal distance of fly =1.587m+1.2m

∴  Horizontal distance of fly =

PLACES FOLLOW ME

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.)