The trajectory of a projectile near the surface of the earth is given as y=2x-9x^{2}. If it were launched at an angle \theta_{0} with speed v_{0} then \left ( g=10ms^{-2} \right ):

  • Option 1)

    \theta_{0}=\sin^{-1}\left ( \frac{2}{\sqrt{5}} \right ) and v_{0}=\frac{3}{5}ms^{-1}

  • Option 2)

    \theta_{0}=\cos^{-1}\left ( \frac{2}{\sqrt{5}} \right ) and v_{0}=\frac{3}{5}ms^{-1}

  • Option 3)

    \theta_{0}=\cos^{-1}\left ( \frac{1}{\sqrt{5}} \right ) and v_{0}=\frac{5}{3}ms^{-1}

  • Option 4)

    \theta_{0}=\sin^{-1}\left ( \frac{1}{\sqrt{5}} \right ) and v_{0}=\frac{5}{3}ms^{-1}

 

Answers (1)

 

Equation of path of a projectile -

y= x \tan \theta\: - \: \frac{gx^{2}}{2u^{2}\cos ^{2}\theta }

it is equation of parabola

g\rightarrow    Acceleratio due to gravity

u\rightarrow    initial velocity

\theta = Angle of projection

 

- wherein

Path followed by a projectile is parabolic is nature.

 

 

Given,      y=2x-9x^{2}

General Eqn.  \rightarrow y= x\tan\theta-\frac{1}{2}\frac{gx^{2}}{v^{2}\cos^{2}\theta}

\Rightarrow \tan\theta=2

\Rightarrow \theta =\sin^{-1}\left ( \frac{2}{\sqrt{5}} \right )=\cos^{-1}\left ( \frac{1}{\sqrt{5}} \right )

\frac{g}{2v^{2}\cos^{2}\theta}=9

\Rightarrow \frac{g}{2v^{2}\times\frac{1}{5}}=9

\Rightarrow \frac{10\times5}{2\times9}=v^{2}

\Rightarrow v^{2}=\frac{25}{9}

v=\frac{5}{3}m/s


Option 1)

\theta_{0}=\sin^{-1}\left ( \frac{2}{\sqrt{5}} \right ) and v_{0}=\frac{3}{5}ms^{-1}

Option 2)

\theta_{0}=\cos^{-1}\left ( \frac{2}{\sqrt{5}} \right ) and v_{0}=\frac{3}{5}ms^{-1}

Option 3)

\theta_{0}=\cos^{-1}\left ( \frac{1}{\sqrt{5}} \right ) and v_{0}=\frac{5}{3}ms^{-1}

Option 4)

\theta_{0}=\sin^{-1}\left ( \frac{1}{\sqrt{5}} \right ) and v_{0}=\frac{5}{3}ms^{-1}

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