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, ho 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?
Answers
Answer:
Step-by-step explanation:
Answer:
2.79 m
Step-by-step explanation:
At first we will have to find the length of AC using pythgoras theorem.
AC2 = (2.4)2 + (1.8)2
AC2 = 5.76 + 3.24 = 9.00AC2
= 5.76 + 3.24 = 9.00
AC = 3 m
length of the string that nazima has out is = 3 m
Given that nazima has pulled the string 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
Now the length of PB
PB2 = PC2 – BC2
= (2.4)2 – (1.8)2
= 5.76 – 3.24 = 2.52
PB = = 1.59 (approx.)
Hence, after 12 seconds the total horizontal distance of the fly from Nazima is
= 1.59 + 1.2 = 2.79 m (approx.)
Answer:
At first we will have to find the length of AC using pythgoras theorem.
AC2 = (2.4)2 + (1.8)2
AC2 = 5.76 + 3.24 = 9.00AC2
= 5.76 + 3.24 = 9.00
AC = 3 m
length of the string that nazima has out is = 3 m
Given that nazima has pulled the string 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
Now the length of PB
PB2 = PC2 – BC2
= (2.4)2 – (1.8)2
= 5.76 – 3.24 = 2.52
PB = = 1.59 (approx.)
Hence, after 12 seconds the total horizontal distance of the fly from Nazima is
= 1.59 + 1.2 = 2.79 m (approx)