A particle is projected at `t = 0` with velocity u at angle `theta` with the horizontal. Then the ratio of the tangential acceleration and the radius of curvature at the point of projection is :

Options

A. `(g^(2)sin2 theta)/(u^(2))`
B. `(2g^(2)sin theta)/(u^(2))`
C. `(g^(2)sin theta)/(u^(2))`
D. `(g^(2)sin 2 theta)/(2u^(2))`

Answer

D

Explanation

R.O.C. `=(u^(2))/(g sin theta)`
`a_(t)=g cos theta`
so `(a_(t))/(R.O.C)=(g cos theta)/(u^(2))xxg`