Question

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, how much string does she have out (see Fig.)? 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?

Answer 1

Let AB is the height of the tip of the fishing rod from the water surface .Let BC is the horizontal distance of the fly from the tip of the fishing rod.

Then AC is the length of the string.

Then according to the Pythagorean theorem-

$\Rightarrow A{C}^2=A{B}^2+B{C}^2$

$\Rightarrow A{C}^2=(1.8{)}^2+(2.4{)}^2$

$\Rightarrow A{C}^2=3.24+5.76$

$\Rightarrow A{C}^2=9.00$

$\Rightarrow AC=\sqrt{9}m=3m$

$\therefore$The length of the string out is $3$ m.

She pulls the string at the rate of 5 cm per second .

$\therefore$She pulls in 12 seconds=$12\times 5=60cm=0.6m$

Let the fly be at a point of D after 12 seconds

Length of the string out of 12 second is AD.

AD=AC-string pull by Nazima after 12 sec.

$AD=3-.6=2.4m$

In $△ADB$

$A{D}^2=A{B}^2+B{D}^2$

$\Rightarrow B{D}^2=A{D}^2-A{B}^2$

$\Rightarrow B{D}^2=(2.4{)}^2-(1.8{)}^2$

$\Rightarrow B{D}^2=5.76-3.24$

$\Rightarrow B{D}^2=2.52$

$\Rightarrow BD=\sqrt{2.52}=1.587m$

Horizontal distance to fly$=BD+1.2$

$\Rightarrow 1.587+1.2\Rightarrow 2.787$ m