A particle (m = 1kg) slides down a frictionless track (AOC) starting from rest at a point A (height 2m). After reaching C, the particle continues to move freely in air as a projectile. When it reaches point C (height 1 m), the vertical speed of the particle (in m/s) is (Figure drawn is schematic and not to scale , take g = 9.8 m/`sec^(2)`) [Take ]

Answer

2.23

Explanation

`mgh_(A) = mgh_(c ) + (1)/(2) mv^(2)`
`V = sqrt(2 g(2 - 1)) = v = 2 sqrt(5) = 2 xx 2.3 = 4.46 m//s`
`V_(y) = V sin 30^(@)`
`V_(y) = v sin 30^(@) = 4.46 xx (1)/(2) = 2.23 m//s`