A particle (m = 2 kg) slides down a smooth track AOC starting from rest at point A (height = 5m). After reaching point C (height =1m), the particle continues to move freely in air as a projectile. The particle velocity at point C is at angle `60^(@)` with the horizontal. The maximum height attained by particle after leaving point C is _____(in m). [figure drawn is schematic and hot to scale : take `=ms^(-2)`]

Answer

C

Explanation

At point C speed of particle is u
`therfore (1)/(2)m v^(2)=mg " "h" "implies (1)/(2)xx2xxu^(2)=2xxx10xx(5-1) implies u^(2)=80`
`therefore h_(max)(u^(2)sin^(2)60^(@))/(2g)=(80xx(3)/(4))/(20)=3m`