A small object slides without friction from the height `5R//2` and then loops of the verticle loop of radius `R` from which a symmetrical section of angle `2alpha` has been removed. Find angle `alpha` such that after losing contact at `A` and flying through the air, the object will reach point `B`.

Explanation

When the block reaches at `P`, its velocity
`u=sqrt(2gh)=sqrt(2gxx(5R)/(2))=sqrt(5gR)`
Velocity at the lowest point `gesqrt(5gR)`, hence the block will complete vertical circle. But at `A`, there is cut, hence it will leave vertical circle due to the cut at `A`.
Let the velocity at `A` be `v`.
`V^(2)=u^(2)+2gh=0+2g[(5R)/(2)-R(1+cosalpha)]`
`=2g((3R)/(2)-R cos alpha)`
`A` to`B` (projectile motion)
Range=`(v^(2)sin2alpha)/(g)`
`2R sin alpha=(2g[(3R)/(2)-R cos alpha]sin 2alpha)/(g)`
`sin alpha=((3)/(2)-cosalpha)(2sin alpha cos alpha)`
`1=(3-2 cos alpha)(cosalpha)`
`2 cos^(2)alpha-3cosalpha+1=0`
`cos alpha=(3+-sqrt(9-8))/(4)=(3+-1)/(4)`
`cos alpha=1impliesalpha=0^(@)`, not possible
`cos alpha=(1)/(2)impliesalpha=60^(@)`