Correct Answer - Option 3 :

\({\theta _0} = {\rm{co}}{{\rm{s}}^{ - 1}}\left( {\frac{1}{{\sqrt 5 }}} \right){\rm{\;and\;}}{v_0} = \frac{5}{3}{\rm{m}}{{\rm{s}}^{ - 1}}\)
**Concept:**

The trajectory of a projectile is given by the equation:

Y = 2x – 9x^{2}

Now, the general equation of trajectory is given by the equation:

\(y = x\;{\rm{tan\;}}\theta - \frac{{g{x^2}}}{{2v_0^2{\rm{co}}{{\rm{s}}^2}\theta }}\)

\( \Rightarrow y = \left( {\tan \theta } \right)x - \left( {\frac{g}{{2v_0^2{\rm{co}}{{\rm{s}}^2}\theta }}} \right){x^2}\)

Where,

y = vertical position (m)

x = horizontal position (m)

v_{0} = initial velocity (combined components, m/s)

g = acceleration due to gravity (m/s^{2})

θ = angle of the initial velocity from the horizontal plane (radians or degrees)

**Calculation:**

Now, comparing given equation with general equation,

The value of the term is obtained.

⇒ tanθ = 2

Now, we need to consider the options,

The tan θ and sin θ relationship is given by the formula:

\(\sin \theta = \frac{{{\rm{tan}}\theta }}{{\sqrt {1 + {\rm{ta}}{{\rm{n}}^2}\theta } }}\)

\( \Rightarrow \sin \theta = \frac{2}{{\sqrt {1 + 4} }}\)

\( \Rightarrow \sin \theta = \frac{2}{{\sqrt 5 }}\)

The tan θ and cos θ relationship is given by the formula:

\(\cos \theta = \frac{1}{{\sqrt {1 + {\rm{ta}}{{\rm{n}}^2}\theta } }}\)

\( \Rightarrow \cos \theta = \frac{1}{{\sqrt {1 + 4} }}\)

\( \Rightarrow \cos \theta = \frac{1}{{\sqrt 5 }}\)

Thus, considering the options, all the options are suitable. So, we need to find the initial speed/velocity value.

Again, comparing given equation with general equation,

\(\frac{g}{{2v_0^2{\rm{co}}{{\rm{s}}^2}\theta }} = 9\)

On substituting the value of cos θ,

\( \Rightarrow \frac{{10}}{{2v_0^2{{\left( {\frac{1}{{\sqrt 5 }}} \right)}^2}}} = 9\)

\( \Rightarrow \frac{{10}}{{2\left( {\frac{1}{5}} \right) \times 9}} = v_0^2\)

\( \Rightarrow \frac{{10 \times 5}}{{2 \times 9}} = v_0^2\)

\( \Rightarrow \frac{{5 \times 5}}{9} = v_0^2\)

\( \Rightarrow \frac{{25}}{9} = v_0^2\)

On taking square root on both sides,

\(\therefore {v_0} = \frac{5}{3}\)

Since, \(\cos \theta = \frac{1}{{\sqrt 5 }}\), the value of initial velocity is \(\frac{5}{3}\).

Therefore, \(\theta = {\cos ^{ - 1}}\left( {\frac{1}{{\sqrt 5 }}} \right)\)

Now, considering the options, option (C) is the correct answer.