Question

A particle is projected from the bottom of an inclined plane of inclination 30^@. At what angle alpha (from the horizontal) should the particle be projected to get the maximum range on the inclined plane.

Answer 1

To get  the maximum range on the inclined plane the particle should be projected at 60°

• Let us go through the given data

• Angle of projection (β)= 30°      

• In the problem they have mentioned that the particle is projected from an inclined plane

       Range = (v²/gcos²β)[(sin(2α-β) - sinβ)

• For the range to be maximum

              α = (π/4) + (β/2)

• By substituting the values

              α = (π/4) + (30/2)

               α = 45° + 15°

               α =60°

• The angle required is 60°