Question

a solider has to cross 80m wide river on a ropeway grabbing on a pulley. the lengt of rope is 100m, its ends are ficed at the same height on both banks fo the river nd ropeis tight. solider stars from under one fixed point with zero initial velo, neglect friction. find highest speed and max depth during the journey

Answer 1

• Maximum speed: $v_{\max}=\sqrt{24\,g}$ (approximately $15.3$ m/s for $g=9.8\,\text{m/s}^2$).
• Maximum depth: $30$ m below the bank level.

Answer 2

We model the situation by taking the two fixed points at A = (0, 0) and B = (80, 0) (the bank level). The soldier, who “grabs” the rope, is at a point (x, y) with y negative (we take downward as negative). The rope is made of two straight‐line segments: one from A to (x, y) and the other from B to (x, y). Since the rope is 100 m long the constraint is

$$\sqrt{x^2+y^2}+\sqrt{(80-x)^2+y^2}=100.$$

Step 1. Find the initial position

The soldier starts “under one fixed point”. Taking this as directly below A means we have $x=0$. Then

$$\sqrt{0^2+y^2} + \sqrt{80^2+y^2} = |y| + \sqrt{6400+y^2} =100.$$

Let $d = |y|$ (with $y=-d$); then

$$d+\sqrt{6400+d^2}=100.$$

Solving for $d$:

$$\sqrt{6400+d^2}=100-d.$$

Square both sides:

$$6400+d^2=(100-d)^2=10000-200d+d^2.$$

Cancel $d^2$ from both sides to get

$$6400=10000-200d\quad\Longrightarrow\quad200d=10000-6400=3600,\quad d=18.$$

Thus the initial position is $(0, y)$ with $y=-18$; that is, the soldier begins 18 m below the level of the banks.

Step 2. Find the lowest (deepest) point

At the lowest point the soldier reaches a minimum $y$ (i.e. maximum drop). We now minimize $y$ subject to the constraint

$$f(x,y)=\sqrt{x^2+y^2}+\sqrt{(80-x)^2+y^2}=100.$$

Use a Lagrange multiplier technique to minimize $y$. Define

$$A=\sqrt{x^2+y^2},\quad B=\sqrt{(80-x)^2+y^2}.$$

Since we are minimizing $y$ (or maximizing $|y|$ with $y<0$), the Lagrange condition tells us that at the extremum the level curves of $y$ are tangent to the constraint curve. Writing the condition we get (details omitted for brevity) that

$$\frac{x}{A} = \frac{80-x}{B}.$$

A little algebra (squaring both sides and cancelling common factors) shows that, for $y\ne0$, one must have

$$x = 80-x \quad\Longrightarrow\quad x=40.$$

Substitute $x=40$ into the constraint:

$$\sqrt{40^2+y^2}+\sqrt{40^2+y^2}=2\sqrt{1600+y^2}=100.$$

Thus,

$$\sqrt{1600+y^2}=50,$$

which gives

$$1600+y^2=2500,\quad y^2=900,\quad y=-30 \quad (\text{since } y<0).$$

So the maximum (absolute) depth is 30 m below the bank level.

Step 3. Find the maximum speed

Since no friction is present the work done by gravity is converted into kinetic energy. Choosing the bank height as zero potential, the gravitational potential energy at any point is $mg\,y$ (with $y<0$). Energy conservation between the starting point ($y=-18$, zero initial speed) and any later position ($y$) gives

$$\frac{1}{2}mv^2 + mg\,y = mg(-18).$$

Thus,

$$\frac{1}{2}v^2= g( -18 - y).$$

The maximum speed occurs when the drop $(-18-y)$ is greatest, that is at the lowest point $y=-30$. Then

$$\frac{1}{2}v_{\max}^2 = g( -18 - (-30))=g(12).$$

So

$$v_{\max}=\sqrt{24\,g}.$$

For example, taking $g=9.8\,\text{m/s}^2$ gives approximately

$$v_{\max}\approx\sqrt{235.2}\approx15.3\,\text{m/s}.$$

Core Explanation Summary:

1. The soldier starts at $(0, y)$ with $y=-18$ since $|y|+\sqrt{6400+y^2}=100$ gives $|y|=18$.
2. Minimizing $y$ with the constraint leads to $x=40$ and $y=-30$ (maximum drop).
3. Conservation of energy from $y=-18$ to $y=-30$ yields $v_{\max}=\sqrt{2g(12)}=\sqrt{24g}$.