In a region, potential energy varies with X as `U(x)=30-(x-5)^(2)` Joule, where x is in meters. A particle of mass 0.5 kg is projected from x=11 m towards origin with a velocity 'u' . u is the minimum velocity, so that the particle can reach the origin. (x=0). find the value of u/2 in meter/second. (Take `sqrt(44)=6.5`)

Explanation

Draw U v/s x graph. There is a maxima of potential energy between x = 11 to x = 0. So to bring the particle from x = 11 to x = 0, the particle has to cross the maxima (x = 5) and to just cross the point x = 5, velocity at x = 5 should be `0^(+)`

`rArr ` Applying energy conservation between x=11 to x=5
`k_(i)+U_(i)=k_(f)+U_(f)`
`1/2(0.5)u^(2)+(30-(11-5)^(2))=0^(+)+(30-(5-5)^(2))`
`u=12 m//sec rArr u/2 = 6m//sec`