Question

A tab AB is fitted to the ground. The tab can throw water in all directions on the ground but the angle \[(\theta )\]that it makes with the horizontal cannot be less than \[30{}^\circ \]and cannot be more than \[60{}^\circ \] also. The initial speed of the water from the tab is \[\sqrt{5}{m}/{s}\;\]. The area of the ground that can be watered is \[\left( \dfrac{\pi }{x} \right){{m}^{2}}\]. What is the value of x? Neglect the length of the tab.

Answer 1

Hint: The maximum range of the projectile formed by the water flow can be calculated by measuring the average value of the ranges of the angles given. In this case, the range of the angles is \[30{}^\circ \] to \[60{}^\circ \]. So, the average will be \[45{}^\circ \]

Formula Used:
\[R=\dfrac{{{u}^{2}}\sin (2\alpha )}{g}\]

Complete step-by-step answer:
From the given, we have the data,
The tab AB is fitted to the ground (not inclined)
The angle made by the water in all direction is \[(\theta )\]
The angle made cannot be less than \[30{}^\circ \]and cannot be more than \[60{}^\circ \].
Thus the angle range made by the water is, \[30{}^\circ <\theta <60{}^\circ \]
The initial speed of the water from the tab is \[\sqrt{5}{m}/{s}\;\].
The area of the ground that can be watered is \[\left( \dfrac{\pi }{x} \right){{m}^{2}}\].

The range of projectile motion is given by the formula,
\[R=\dfrac{{{u}^{2}}\sin (2\alpha )}{g}\]
Where u is the initial speed of the particle, \[\alpha \]is the angle made by the particle with respect to the ground and g is the gravitational constant.

The range of the water thrown at angles \[30{}^\circ \]and \[60{}^\circ \] will be the same, because the value of angle remains the same.
Consider the least value of angle
\[\begin{align}
  & \sin (2\alpha ) \\
 & =\sin (2\times 30{}^\circ ) \\
 & =\sin (60{}^\circ ) \\
 & =\dfrac{\sqrt{3}}{2} \\
\end{align}\]
Consider the highest value of angle
\[\begin{align}
  & \sin (2\alpha ) \\
 & =\sin (2\times 60{}^\circ ) \\
 & =\sin (120{}^\circ ) \\
 & =\dfrac{\sqrt{3}}{2} \\
\end{align}\]

Now compute the range of water thrown at the angles \[30{}^\circ \]and \[60{}^\circ \].
\[\begin{align}
  & {{R}_{1}}=\dfrac{{{u}^{2}}\sin (2{{\alpha }_{1}})}{g} \\
 & {{R}_{1}}=\dfrac{{{(\sqrt{5})}^{2}}\times \sin (2\times 30{}^\circ )}{10}=\dfrac{{{(\sqrt{5})}^{2}}\times \sin (2\times 60{}^\circ )}{10} \\
 & {{R}_{1}}=\dfrac{{{(\sqrt{5})}^{2}}\times \sin 60{}^\circ }{10}=\dfrac{{{(\sqrt{5})}^{2}}\times \sin 120{}^\circ }{10} \\
 & {{R}_{1}}=\dfrac{5}{10}\times \dfrac{\sqrt{3}}{2} \\
 & {{R}_{1}}=\dfrac{\sqrt{3}}{4} \\
\end{align}\]

The range of water thrown will be maximum at an angle of \[45{}^\circ \], as this angle is perpendicular to the tab.
\[\begin{align}
  & {{R}_{2}}=\dfrac{{{u}^{2}}\sin (2{{\alpha }_{2}})}{g} \\
 & {{R}_{2}}=\dfrac{{{(\sqrt{5})}^{2}}\times \sin (2\times 45{}^\circ )}{10} \\
 & {{R}_{2}}=\dfrac{{{(\sqrt{5})}^{2}}\times \sin 90{}^\circ }{10} \\
 & {{R}_{2}}=\dfrac{5}{10}\times 1 \\
 & {{R}_{2}}=\dfrac{1}{2} \\
\end{align}\]

The area covered by the water is given by, \[A=\pi ({{R}_{2}}{}^{2}-{{R}_{1}}{}^{2})\]
Comparing the equation of given area of water \[A=\left( \dfrac{\pi }{x} \right)\]with equation, we get,
\[\begin{align}
  & \left( \dfrac{\pi }{x} \right)=\pi ({{R}_{2}}{}^{2}-{{R}_{1}}{}^{2}) \\
 & \dfrac{1}{x}={{R}_{2}}{}^{2}-{{R}_{1}}{}^{2} \\
\end{align}\]

Substitute the obtained values of the ranges.
\[\begin{align}
  & \dfrac{1}{x}=\left( \dfrac{1}{2} \right){}^{2}-\left( \dfrac{\sqrt{3}}{4} \right){}^{2} \\
 & \dfrac{1}{x}=\left( \dfrac{1}{4} \right)-\left( \dfrac{3}{16} \right) \\
 & \dfrac{1}{x}=\dfrac{1}{16} \\
 & x=16 \\
\end{align}\]

Therefore, the value of x is 16.

Note: The things to be on your finger-tips for further information on solving these types of problems are: The units of the given parameters should be taken into consideration. The horizontal range is the product of horizontal velocity and the time of flight. If any two of these parameters are known, the third parameter can be found out easily.