A particle is projected with a velocity `'9 m//sec'` at an angle `45^(0)`, with the horizontal.What is the radius of curvature of the trajectory of the particle at the position `'x=R//3'` (R-Range of the projectile).

Options

A. `3sqrt20m`
B. `3sqrt10`
C. `(3sqrt10)/2 m`
D. `3/4sqrt10m`

Answer

A, B, C

Explanation

The equation of the trajectory of the projectile is,
`y=x tan theta (1-(x)/(R)).` Here `theta=45^(@)`
Let `'beta'` be the angle made by the instantaneous velocity vector with the horizontal at position `x=R//3`
`m=tan beta=(dy)/(dx)=1-(2x)/R=1//3`
Let `v` be the velocity of the projectile at that instant
`v cos beta=u cos 45 rArr vxx3/sqrt10=uxx1/sqrt2`
`rArr u=(sqrt10xxu)/(3sqrt2)`
`a_(N)=g cos beta=v^(2)/R`
`:.R=v^(2)/(gcos beta)=(10xxu^(2))/(9xx2xx10xx3/sqrt10)`
`=(9xx9)/(9xx2xx3)xxsqrt10=(3sqrt10)/2`