Hard
Question
A plane surface is inclined making an angle θ 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
Options
A.
B.
C.
D.
Solution
On the incline plane the maximum possible Range is -
R =
uy = u sin α ux = cos α ax = − g sin β ay = − g cos β
Range = sx = ux T +
ax T2
(on the inclined plane) where T =
⇒ sx = (u cos α)
=
[cos αcos β − sin αsin β]
sx =
[cos (α + β)]
=
[2sinαcos(α + β)]
=
[sin (2α + β) + sin (−β)]
sx =[sin(2α + β) − sin β]
Now sx 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.
& (sx)max =
= 
Max. Range on an inclined plane =
Here β = θ ⇒ Rmax =
Ans ″B″
R =
uy = u sin α ux = cos α ax = − g sin β ay = − g cos β
Range = sx = ux T +
ax T2(on the inclined plane) where T =
⇒ sx = (u cos α)
=
[cos αcos β − sin αsin β]sx =
[cos (α + β)]=
[2sinαcos(α + β)]=
[sin (2α + β) + sin (−β)]sx =
[sin(2α + β) − sin β]Now sx 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.
& (sx)max =
= Max. Range on an inclined plane =
Here β = θ ⇒ Rmax =
Ans ″B″Create a free account to view solution
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