Q. 22

# Naman is doing fl

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

𝐶𝑀2 = 𝐵𝐶2𝑀𝐵2

C’M2 = (2.4)2− (1.8)2 = 2.52

CM = 1.6 m

The horizontal distance of the fly from him after 12 seconds is given by

CA = CM + MA = 1.6 + 1.2 = 2.8 m

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