Question

Nazima is fly fishing in a stream. The tip of her fishing rod is $1.8m$ above the surface of the water and the fly at the end of the string rests on the water $3.6m$ away and $2.4m$ 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 Fig.? If she pills in the string at the rate of $5cm$ per second, what will be the horizontal distance of the fly from her after $12$ seconds?

Answer 1

$AB=$ height of the tip of the fishing rod from the water surface.
$BC=$ Horizontal distance of the fly from the tip of the sishing rod.
$AC=$ Length of the string.
Step 1: Applying pythagoras theorem in $\Delta ACB$
$A{C}^2=A{B}^2+B{C}^2$ [By pythagoras theorem]
$A{C}^2={1.8}^2+{2.4}^2$
$A{C}^2=3.24+5.76\Rightarrow A{C}^2=9$
$\Rightarrow AC=3m=$ string length

Step 2: Finding out length of string out after 12 second
Rate of pulling in string $=5cm/s$
$\therefore$ Atring pulled in 12 second $=12\times 5=60cm$
$=0.6m$
Position of fly after 12 seconds $=D$.
length of string out after 12 second $=AD$.
$AD=AC-$ String pulled in 12 seconds. (By the girl)
$Ad=3-0.6=2.4m$

Step 3 : Applying pythagoras theorem in $\Delta ADB$ in $\Delta ADB$,
$A{D}^2=A{B}^2+B{D}^2$ [By pythagoras theorem]
${2.4}^2={1.8}^2+B{D}^2$
$B{D}^2=5.76-3.24$
$B{D}^2=2.52$
$\Rightarrow BD=1.587m$
$\Rightarrow$ Horizontal distance of fly $=BD+1.2m$
Horizontal distance of fly $=1.587m+1.2m=2.787m$