Question

A small particle is given an initial velocity `v_(0)=10m//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 h = 15 m below A and a distance r from the vertical centerline, where its velocity v makes an angle `theta` with a the horizontal tangent to the bowl through B. Find the value `18cos^(2)theta` (take g = 10 `m//s^(2)`).

Answer 1

Correct Answer - 2
`t_(DE)=sqrt((2d)/(g))`
`4d=(v_(0))sqrt((2d)/(g))`
`v_(0)^(2)=8gd`
energy - work theorem
`(1)/(2)kdelta^(2)=(mumgd)+mgd+(1)/(2)m(v_(0))^(2)`