A particle is fired from a point on the ground with speed `u` making an angle `theta` with the horizontal.Then

Options

A. the radius of curvature of the projectile at the highest point is `(u^(2)cos^(2) theta)/g`
B. the radius of curvature of the projectile at the highest point is `(u^(2)sin^(2) theta)/g`
C. at the point of projection magnitude of tangential acceleration is `g sin theta`
D. at the point of projection magnitude of tangential acceleration is `g cos theta`

Answer

A, C

Explanation

At hightest point angle between `veca` and `vecv` is zero.Hence, total acceleration is only normal or radial acceleration.
`:.a=a_(n)V^(2)/R :.a=a_(n)V^(2)/R`
but `a=g` but `a=g`
`:. g(u cos theta)^(2)/R :.g(u cos theta)^(2)/R`
or `R=(u^(2)cos^(2)theta)/g `or `R=(u^(2)cos^(2)theta)/g `
At point of projection component of acceleration `(=g)` along velocity vector is `-g cos (90^(@)-theta)` or
`-g sin theta`.