Question

Nazima is doing 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 from her and 2.4 m from the point directly under the tip of the rod. Assuming that the string (from the tip of her rod to the fly) is taut, how much string does she have out (see the adjoining figure)? 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 A be the position of the fishing rod above the water.
Thus, AB = 1.8 m

The distance between the fly from Nazima = 3.6 m

i.e, CD = 3.6 m

Also, CB = 2.4 m

and thus BD = 3.6 - 2.4 = 1.2 m

AC is the length of the string Nazima has out. We need to find AC.

Since AB is a straight line.

$△$ABC is a right-angled triangle,

By Pythagoras theorem,

$A{C}^2=A{B}^2+B{C}^{2}A{C}^2=(1.8{)}^2+(2.4{)}^{2}A{C}^2=3.24+5.76A{C}^2=9AC=3m$

When she pulls the string:

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

Length of string pulled in 12 seconds $=5\times =60cm=0.6m$

Our new figure will be:

So,$AP=AC-0.6=3-0.6=2.4m$

We need to find the horizontal distance of the fly from her. Let this distance be BP,

Now in $△$ABP,

By Pythagoras theorem,

$A{P}^2=A{B}^2+B{P}^{2}B{P}^2=(2.4{)}^2-(1.8{)}^{2}B{P}^2=5.76-3.24B{P}^2=2.52BP=1.58m$

The total horizontal distance of the fly from her
$=DP=BP+BD=1.58+1.2=2.78m$