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  • A plane surface is inclined making an angle  with the horizontal. from the bottom of this inclined plane

A plane surface is inclined making an angle \theta with the horizontal. from the bottom of this inclined plane a bullet is fired with velocity v. The maximum possible range of the bullet on the inclined plane is 

Option: 1

\frac{v^{2}}{g}

 

 

 


Option: 2

\frac{v^{2}}{g\left ( 1+\sin \theta \right )}


Option: 3

\frac{v^{2}}{g\left ( 1-\sin \theta \right )}


Option: 4

\frac{v^{2}}{g\left ( 1+\cos \theta \right )}


Answers (1)

best_answer

For projectile on an inclined plane -

Range on inclined plane up to the plane is given 

R=\frac{u^2}{gcos^2\theta}.2sin(\alpha-\theta)cos\alpha

\Rightarrow R=\frac{u^{2}}{g\cos ^{2}\theta }\left [ \sin \left ( 2\alpha -\theta \right )-\sin \theta \right ]

Where 

\alpha = Angle of projection above inclined plane

        (measured from horizontal line)

 \theta=Angle of inclination.

 

So for maximum range \sin \left ( 2 \alpha -\theta \right ) should be maximum 

so for \left ( 2\alpha - \theta \right )=\frac{\pi }{2}

R=\frac{u^{2}}{g\cos ^{2}\theta }\left [ 1-\sin \theta \right ]

R=\frac{u^{2}}{g\left ( 1-\sin ^{2}\theta \right ) }\left [ 1-\sin \theta \right ]

R=\frac{u^{2}}{g\left ( 1+\sin \theta \right ) }

 

Posted by

vinayak

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