A peacock is sitting on the top of a pillar, which is $9 m$ high. From a point $27 m$ away from the bottom of the pillar a snake is coming to its hole at the base of the pillar. Seeing the snake the peacock pounces on it. If their speeds are equal, at what distance from the whole is the snake caught?

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Solution

Let $P Q$ be the pole and the peacock is sitting at the top $P$ of the pole. Let the hole be at $Q$ . Initially, the snake is at $S$ when the peacock notices the snake such that $Q S = 27 m$

Suppose $v$ m/sec is the common speed of both the snake and the peacock and the peacock catches the snake after $t$ seconds at point $T$ . clearly distance travelled by the snake in $t$ seconds is same as the distance flown by peacock

$∴ P T = S T = x$

Thus, in right triangle $P Q T$ , we have

$Q T = 27 - x, P T = x$ and $P Q = 9$

Using Pythagoras theorem, we have

${P T}^{2} = {P Q}^{2} + {Q T}^{2} \Rightarrow x^{2} = 9^{2} + {(27 - x)}^{2}$

$\Rightarrow x^{2} = 81 + 729 - 54 x + x^{2} \Rightarrow 0 = 810 - 54 x \Rightarrow 54 x = 810 \Rightarrow x = 15$

$∴ Q T = S Q - S T = (27 - 15) m = 12 m$

Hence the snake is caught at a distance of $12 m$ from the hole.

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A peacock is sitting on the top of a pillar, which is 9 m high. From a point 27 m away from the bottom of the pillar a snake is coming to its hole at the base of the pillar. Seeing the snake the peacock pounces on it. If their speeds are equal, at what distance from the whole is the snake caught?