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  • Please,please help me A particle is projected in horizontal direction from a height. Initial speed is 4m/s. Then the angle made by its velocity with horizontal direction after 1 second is

A particle is projected in horizontal direction from a height. Initial speed is 4m/s. Then the angle made by its velocity with horizontal direction after 1 second is 

  • Option 1)

    \tan^{-1}\left(\frac{2}{5}\right)

  • Option 2)

    \tan^{-1}\left(\frac{5}{2}\right)

  • Option 3)

    \tan^{-1}\left(\frac{1}{2}\right)

  • Option 4)

    \tan^{-1}\left(\frac{1}{3}\right)

 

Answers (1)

best_answer

As we have learnt,

 

Projectile Projected Horizontally -

\tan \beta = \frac{gt}{u}

 

- wherein

\beta =  angle that velocity makes with horizontal

 

 Angle with horizontal is

\theta=\tan^{-1}\left (\sqrt{\frac{y}{x}} \right ) = \tan^{-1}\left(\frac{gt}{4x} \right )\Rightarrow \theta = \tan^{-1} (\frac{10}{4}) = \tan^{-1}(\frac{5}{2})

 


Option 1)

\tan^{-1}\left(\frac{2}{5}\right)

Option 2)

\tan^{-1}\left(\frac{5}{2}\right)

Option 3)

\tan^{-1}\left(\frac{1}{2}\right)

Option 4)

\tan^{-1}\left(\frac{1}{3}\right)

Posted by

Avinash

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