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?

Question

Philoid · Accepted Answer

Naman pulls in the string at the rate of 5 cm per second.

Hence, after 12 seconds the length of the string he will pull is given by

12 × 5 = 60 cm or 0.6 m

Now, in ΔBMC

By using Pythagoras theorem, we have

BC² = CM² + MB²

⇒ BC² = (2.4)² + (1.8)² = 9

∴ BC = 3 m

Now, BC’ = BC – 0.6 = 3 – 0.6 = 2.4 m

Now, in ΔBC’M

By using Pythagoras theorem, we have