Understanding the Problem: Peacock, Snake, and Distance

The question describes a scenario involving a peacock on top of a pole and a snake approaching its base. Both animals move at the same speed, and we need to find the distance from the pole where the peacock catches the snake after jumping from the pole.

Setting Up the Scenario

Let's break down the information given:

• Height of the pole: $h = 3$ meters.
• Initial distance of the snake from the pole: This is three times the height of the pole. Distance $d = 3 \times h = 3 \times 3 = 9$ meters.
• Speed of the peacock = Speed of the snake. Let this speed be $v$.
• The peacock jumps from the top of the pole and catches the snake. We assume the peacock flies in a straight line from the top of the pole to the point of capture on the ground.

We need to find the distance from the base of the pole where the catch occurs.

Analyzing the Movement and Distances

Let's denote the point at the base of the pole as B, the top of the pole as T, and the initial position of the snake as S. Let the point where the peacock catches the snake be C. The point C will be on the line BS (the snake's path towards the pole).

The initial distance BS is 9 meters. Let the distance from the pole base B to the point of capture C be $y$ meters. This is the distance we want to find.

The snake starts at S and moves towards B, reaching C. The distance the snake travels is SC. Since B is between S and C (or C is between S and B), and BC is $y$, the distance the snake travels is $SC = BS - BC = 9 - y$ meters.

The peacock starts at T and moves towards C. The path of the peacock is the line segment TC. The height BT is 3 meters. Triangle TBC is a right-angled triangle with the right angle at B.

Using the Pythagorean theorem in triangle TBC, the distance the peacock travels (TC) is given by:

$\text{TC}^2 = \text{TB}^2 + \text{BC}^2$

$\text{TC}^2 = h^2 + y^2$

$\text{TC} = \sqrt{h^2 + y^2} = \sqrt{3^2 + y^2} = \sqrt{9 + y^2}$ meters.

Relating Distances with Speed and Time

Both the peacock and the snake move for the same amount of time, let's call it $t$, from the moment the peacock sees the snake until it catches it. Since they move at the same speed $v$, the distance covered by each must be equal.

Distance covered by snake (SC) = $v \times t$

Distance covered by peacock (TC) = $v \times t$

Therefore, $\text{SC} = \text{TC}$.

Substituting the distances we found:

$9 - y = \sqrt{9 + y^2}$

Solving for the Distance from the Pole

We need to solve the equation $9 - y = \sqrt{9 + y^2}$ for $y$.

Square both sides of the equation to eliminate the square root:

$(9 - y)^2 = (\sqrt{9 + y^2})^2$

$(9 - y)(9 - y) = 9 + y^2$

Expand the left side (using $(a-b)^2 = a^2 - 2ab + b^2$):

$9^2 - 2(9)(y) + y^2 = 9 + y^2$

$81 - 18y + y^2 = 9 + y^2$

Subtract $y^2$ from both sides of the equation:

$81 - 18y = 9$

Subtract 9 from both sides:

$81 - 9 = 18y$

$72 = 18y$

Divide by 18 to find $y$:

$y = \frac{72}{18}$

$y = 4$

Conclusion

The distance from the base of the pole where the peacock catches the snake is 4 meters.