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The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectiles is:

  • Option 1)

    \theta= \tan^{-1}(\frac{1}{4})

  • Option 2)

    \theta= \tan^{-1}(4)

  • Option 3)

    \theta= tan^{-1}(2)

  • Option 4)

    \theta= 45^{0}

 

Answers (1)

As we learnt in

Horizontal Range -

Horizontal distance travelled by projectile from the point of projectile to the point on ground where its hits.

R=frac{u^{2}sin 2Theta }{g}

 

 

 

- wherein

Special case of horizontal range

For man horizontal range.

Theta = 45^{0}

R_{max}=frac{u^{2}sin 2 (45) }{g}=frac{u^{2}	imes 1}{g}=frac{u^{2}}{g}

 

 

 

 

 

R=H

\frac{u^{2}sin{2}\theta}{g}\:=\:\frac{u^{2}sin^{2}\theta}{2g}

tan\theta=4\:\:\:\:\:\:\:\:\:=> \theta=tan^{-1}4

Correct option is 2.


Option 1)

\theta= \tan^{-1}(\frac{1}{4})

Incorrect

Option 2)

\theta= \tan^{-1}(4)

Correct

Option 3)

\theta= tan^{-1}(2)

Incorrect

Option 4)

\theta= 45^{0}

Incorrect

Posted by

Vakul

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