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A gardener wants to wet a garden without moving from his place with a water whose velocity is $20\,m{{s}^{-1}}$. What is the maximum area (in ${{m}^{2}}$) that he can wet? ($g=10\,m{{s}^{-2}}$)
A. $1600\pi $
B. $40\pi $
C. $400\pi $
D. $200\pi $

Last updated date: 13th Jun 2024
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Hint: This question is related to projectile motion. The object projected from a place near earth’s surface cannot cover a horizontal distance more than a certain limit known as its maximum horizontal range. The maximum area that can be covered will be a circular area of radius equal to maximum horizontal range. We need to calculate the area of the circle.

Formula used: Maximum range of a projectile, $R=\dfrac{{{u}^{2}}}{g}$; Area of a circle, $A=\pi {{r}^{2}}$

Complete step by step answer:
When an object is projected near the earth’s surface, it moves along a parabolic path under constant acceleration due to gravity. This motion is known as projectile motion. The path followed by the object is known as its trajectory.
When an object is projected with initial velocity $u$, its maximum horizontal range is given by
The angle of projection for maximum range must be ${{45}^{{}^\circ }}$.
When the gardener projects the water from his position, he can cover a circular area of maximum radius equal to the range of the projectile. That is
The area of circle is given by
$A=\pi {{r}^{2}}$
We substitute $r=R=\dfrac{{{u}^{2}}}{g}$ and get maximum area the gardener can water without moving from his place
$A=\pi {{R}^{2}}=\pi {{\left( \dfrac{{{u}^{2}}}{g} \right)}^{2}}$
Since velocity of projection of water is $20m/s$, we substitute $u=20m/s$ and $g=10m/{{s}^{2}}$
$A=\pi {{\left( \dfrac{{{20}^{2}}}{10} \right)}^{2}}=1600\pi \,{{m}^{2}}$
The maximum area (in ${{m}^{2}}$) of the garden that he can wet without moving from his place is $1600\pi $.

So, the correct answer is “Option A”.

Note: To wet the maximum area the gardener must be inside the garden, at least a distance equal to maximum range, from each side of the garden.
When an object is projected at an angle $\theta ={{45}^{{}^\circ }}$ with horizontal, it will cover maximum horizontal range.
To cover maximum height, the object must be projected at an angle $\theta=90^\circ$ with horizontal.