A plane surface is inclined making an angle θ with the horizontal. From the bottom of this incl

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A plane surface is inclined making an angle &#952; 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

Accepted Answer

On the incline plane the maximum possible Range is -
R =

u_y = u sin α     u_x = cos α     a_x = − g sin β     a_y = − g cos β
Range = s_x = u_x T + a_x T²
(on the inclined plane) where T =
⇒ s_x = (u cos α)

= [cos αcos β − sin αsin β]
s_x = [cos (α + β)]
= [2sinαcos(α + β)]
=[sin (2α + β) + sin (−β)]
s_x = [sin(2α + β) − sin β]
Now s_x is max
when sin (2α + β) is max        (∵ β = constt)
⇒    2α + β =      ⇒
i.e., when ball is projected at the angle bisector of angle formed by inclined plane and dir. of net acceleration reversed.
& (s_x)_max = =
Max. Range on an inclined plane =

Here β = θ ⇒ R_max = Ans ″B″

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