Chapter 11.6, Problem 12P

To determine

Tofind:the distance that he moves the boat.

Answer to Problem 12P

  $1.5m$ .

Explanation of Solution

Given:

The height of the dock from the water surface is $2.0\text{m}$ .

Gary is pulling the boat on the dock from the rope of length is $5.2\text{m}$ .

Calculation:

According to the question, Garypulls the rope of the boat is $2.3m$ .

Now, draw the sketch of the above condition:

  

Now, from Pythagoras theorem, $A{C}^2=A{B}^2+B{C}^2$ .

Therefore, the length of each diagonal of the rectangular box is

  $\begin{array}{l}A{C}^2=A{B}^2+B{C}^2\text{ {}\because AB=2.0,AC=5.2\text{}}\\ {\left(5.2m\right)}^2={\left(2.0m\right)}^2+B{C}^2\\ 27.04m^2=4m^2+B{C}^2\\ \therefore B{C}^2=27.04m^2-4m^2\\ B{C}^2=23.04m^2\\ BC=\sqrt{23.04m^2}\text{{Taking square both sides}}\\ BC=4.8m\end{array}$ .

Thus, the distance that he moves the boat is $4.8-2.3=1.5m$ .

Hence, the distance that he moves the boatis $1.5m$ .