A particle is projected up the inclined such that its component of velocity along the incline is `10 m//s`. Time of flight is `2` sec and maximum height above the incline is `5m`. Then velocity of projection will be

Options

A. `10 m//s`
B. `10 sqrt(2) m//s`
C. `5sqrt(5) m//s`
D. none

Answer

B

Explanation

Using the given data in the formulae for projection up the inclined plane.
(`theta` is angle of projectile with inclined plane, `beta` is angle of inclined with horizontal)
`T=2 sec =(2v sin theta)/(g cos beta)`
`v cos theta=10.....(1)`
`h=5 =(v sin theta)/2xx1`
`v sin theta =10`
from (i) and (ii)
`v^(2)=200 `
`v=10sqrt(2) m//s`