A particle is projected with speed `v = sqrt(150)m//s` from the horizontal surface such that its range on the horizontal plane is twice the greatest height attained by it. Then half of range of the projectile in meter is (use g = 10 `m//s^(2)`)

Answer

3

Explanation

`v=sqrt(2(10)Rcos60^(@))=sqrt(10)`
`N-mgcos60^(@)=(mv^(2))/(R )implies N=15` Newton
`therefore" Force on wall = N sin "60^(@)=(15sqrt(3))/(2) " "therefore P = 3`