# The trajectory of a projectile near the surface of the earth is given as y = 2x - 9x2. If it were launched at an angle θ0 with speed v0 then (g = 10 m

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The trajectory of a projectile near the surface of the earth is given as y = 2x - 9x2. If it were launched at an angle θ0 with speed v0 then (g = 10 ms-2):
1. ${\theta _0} = {\rm{si}}{{\rm{n}}^{ - 1}}\left( {\frac{1}{{\sqrt 5 }}} \right){\rm{\;and\;}}{v_0} = \frac{5}{3}{\rm{m}}{{\rm{s}}^{ - 1}}$
2. ${\theta _0} = {\rm{co}}{{\rm{s}}^{ - 1}}\left( {\frac{2}{{\sqrt 5 }}} \right){\rm{\;and\;}}{v_0} = \frac{3}{5}{\rm{m}}{{\rm{s}}^{ - 1}}$
3. ${\theta _0} = {\rm{co}}{{\rm{s}}^{ - 1}}\left( {\frac{1}{{\sqrt 5 }}} \right){\rm{\;and\;}}{v_0} = \frac{5}{3}{\rm{m}}{{\rm{s}}^{ - 1}}$
4. ${\theta _0} = {\rm{si}}{{\rm{n}}^{ - 1}}\left( {\frac{2}{{\sqrt 5 }}} \right){\rm{\;and\;}}{v_0} = \frac{3}{5}{\rm{m}}{{\rm{s}}^{ - 1}}$

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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 – 9x2

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)

v0 = initial velocity (combined components, m/s)

g = acceleration due to gravity (m/s2)

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.