Question

Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m about 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 given figure)? If she pulls the string at the rate of 5 cm per second, what will the horizontal distance of the fly from her after 12 second?

Answer 1

Let $AB$ 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.

$\therefore$ $A{C}^2=A{B}^2+B{C}^2$ [ By Pythagoras theorem ]

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

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

$\therefore$ $A{C}^2=9$

$\therefore$ $AC=3m$

Thus, the length of the string out is $3m$

She pulls the string at the rate of $5cm/s$

$\therefore$ String pulled in $12$ second = $12\times 5=60cm=0.6m$

Let the fly be at point $D$ after 12 seconds.

Length of string out after 12 second is $AD$.

$\Rightarrow$ AD = AC - String pulled by Nazima in 12 seconds.

$\Rightarrow$ $AD=(3-0.6)m=2.4m$

$\Rightarrow$ In $△ADB$,

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

$(1.8{)}^2+B{D}^2=(2.4{)}^2$

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

$B{D}^2=2.52$

$\therefore$ $BD=1.587m$

Horizontal distance of fly $=BD+1.2m$

Horizontal distance of fly $=1.587m+1.2m$

$\therefore$ Horizontal distance of fly $=2.79m$