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A projectile is fired at an angle of 45^{o} with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection is 

  • Option 1)

    60^{o}

  • Option 2)

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

  • Option 3)

    \tan ^{-1}\left ( \frac{\sqrt{3}}{2} \right )

  • Option 4)

    45^{o}

 

Answers (1)

Using

Maximum Height -

Maximum vertical distance attained by projectile during its journey.

H= frac{U^{2}sin ^{2}Theta }{2g}

- wherein

Special Case

If U is doubled , H become four times and T becomes twice provided Q & g are constant.

 

 and

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}

 

 H = \frac{{v_{0}}^{2}\sin ^{2}45^{\circ}}{2g} = \frac{{v_{0}}^{2}}{4g}\ \ \ -(i)

R = \frac{{v_{0}}^{2}\sin 90}{g} = \frac{{v_{0}}^{2}}{g}

\frac{R}{2} = \frac{{v_{0}}^{2}}{2g}\ \ \ -(ii)

\therefore \tan \alpha = \frac{H}{\frac{R}{2}} = \frac{{v_{0}}^{2}}{4g}\times \frac{2g}{{v_{0}}^{2}} = \tan \alpha

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


Option 1)

60^{o}

This option is incorrect

Option 2)

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

This option is correct

Option 3)

\tan ^{-1}\left ( \frac{\sqrt{3}}{2} \right )

This option is incorrect

Option 4)

45^{o}

This option is incorrect

Posted by

subam

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