A particle is projected along an inclined plane as shown in figure. What is the speed of the particle when it collides at point A ? `(g = 10 m/s^2)`

Answer

A, C

Explanation

`T = (2usin(alpha-beta))/(gcosbeta)`
`=(2xx10xxsin30^@)/((10)cos30^@)`

`=2/(sqrt3)`s
`v_x = u_x = 10 cos 60^@ = 5 m//s`
`v_y = u_y + a_yt`
`=(10sin60^@) + (-10) (2sqrt(3))`
`=5(sqrt3) - 20/(sqrt3) = -5/(sqrt3) m//s`
`=v=(sqrt(v_(x)^2 + v_(y)^2))`
`= (sqrt(25 + (25)/3))`
`=10/(sqrt3) m//s`.