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A plane surface is inclined making at 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
$\frac{v^{2}}{g(1+\sin \theta)}$

Option: 3
$\frac{v^{2}}{g(1-\sin \theta)}$

Option: 4
$\frac{v^{2}}{g(1+ \cos \theta)}$

Answers (1)

best_answer

Projectile on an inclined plane -

Important equations

U=Speed of projection

$\alpha=$ Angle of projection above inclined plane (measured from horizontal line)

$\theta =$ Angle of projection above inclined plane (measured from inclined plane)

$\beta =$ Angle of inclination.

a) Initial Velocity- U

Component along x or along inclined plane = $U_{x} = Ucos\theta$

Component along y or perpendicular to inclined plane = $U_{y} = USin\theta$

b) Final velocity =V

Component along x or along inclined plane = $V_{x} = Ucos\theta - (gsin\beta).t$

Component along y or perpendicular to inclined plane = $V_{y} = Usin\theta - (gcos\beta).t$

and, $V = \sqrt{V_{x}^{2} + V_{y}^{2} }$

c) Displacement=S

Component along x or along inclined plane = $S_{x} = U_{x}t + \frac{1}{2}a_{x}.t^{2}$

Component along y or perpendicular to inclined plane = $S_{y} = U_{y}t + \frac{1}{2}a_{y}.t^{2}$

And $S = \sqrt{S_{x}^{2} + S_{y}^{2} }$

d) Acceleration = a

Component along x or along inclined plane= $a_{x} = -gsin\beta$

Component along y or perpendicular to inclined plane = $a_{y} = -gcos\beta$

So a=-g

For maximum range $\sin \left ( 2\alpha -\theta \right )$ should be maximum

So for $\left ( 2\alpha -\theta \right )= \frac{x}{2}$

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

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

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

Posted by

shivangi.bhatnagar

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