Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8m above thesurfaceof 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. 7.235)? If she pulls the string at the rate of 5 cm per second, what will the
horizontal distance of the fly from her after 12 seconds .
Answers
Answer:
2.79m
Step-by-step explanation:
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.
∴ AC
2
=AB
2
+BC
2
[ By Pythagoras theorem ]
∴ AC
2
=(1.8)
2
+(2.4)
2
∴ AC
2
=3.24+5.76
∴ AC
2
=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,
AB
2
+BD
2
=AD
2
(1.8)
2
+BD
2
=(2.4)
2
BD
2
=5.76−3.24
BD
2
=2.52
∴ BD=1.587m
Horizontal distance of fly =BD+1.2m
Horizontal distance of fly =1.587m+1.2m
∴ Horizontal distance of fly =2.79m
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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.)