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10. Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surfaceof the water and the fly at the end of thestring rests on the water 3.6 m away andthe rod. Assuming that her string2.4 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 atthe rate of 5 cm per second, what will bethe horizontal distance of the fly from herafter 12 seconds? |
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Answer» ANSWERLet 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=3mThus, the length of the string out is 3mShe pulls the string at the rate of 5cm/s∴ String pulled in 12 second = 12×5=60cm=0.6mLet 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)2BD2=5.76−3.24BD2=2.52∴ BD=1.587mHorizontal distance of fly =BD+1.2mHorizontal distance of fly =1.587m+1.2m∴ Horizontal distance of fly =PLACES FOLLOW ME |
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