A small particle is given an initial velocity `v_(0)=10 m//s` along the tangent to the brim of a fixed smooth hemisphere bowl of radius `r_(0)=15sqrt(2)m` as shown in the figure. The particle slides on the inner surface and reaches point `B`, a vertical distance `h=15m` below `A` and a distance `r` from the vertical centerline, where its velocity `v` makes an angle `theta` with the horizontal tangent to the bowl through `B`. If `theta=15 K^(@)`. Find the value of `K`. (take `g=10 m//s^(2)`)

Explanation

Angular momentum of the particle is conserved about the vertical central line
`mv_(0)r_(0)=mvrcostheta`
where conservation of mechanical energy gives.
`(1)/(2)mv_(0)^(2)+mgh=(1)/(2)mv^(2)`
`r^(2)=r_(0)^(2)-h^(2)`
`costheta=(1)/(sqrt(1+(2gh)/(v_(0)^(2)))sqrt(1-(h^(2))/(r_(0)^(2))))`