A particle is projected with a velocity `8 m//sec` at an angle `45^(0)`, with the horizontal.What is the radius of curvature of the trajectory of the particle at the instant of `'1/4th'` of the time of ascent.

Options

A. `6.25 m`
B. `12.5 m`
C. `8 m`
D. `10 m`

Answer

B

Explanation

The angle made by the velocity vector with the horizontal at `'t=t_(alpha)/4` is:
`tan beta=(v_(y))/(v_(x))=(u_(y)-"gt")/u_(x)=3/4`
Let `v` be the velocity of the particle at `'t=t_(a)/4`:
then `v cos beta=u cos 45`
`vxx4/5=u(1)/sqrt2 rArr v=(5u)/(4sqrt2)m//sec`
Hence normal acceleration `a_(N)=g cos beta =v^(2)/R`
`:. R=v^(2)/ (g cos beta)=(25u^(2))/(16xx2xx10xx4/5)`
`=(25xx8xx8)/(32xx8)=6.25m`