A projectile is fired from the ground at a speed of 20 m/s at an angle of ${37}^{\circ }$ to the horizontal, then $[g=10m/s^2]$

The correct options are
A Radius of curvature at firing point is 50 m.
C As projectile moves its radius of curvature first decreases and then it increases.

By relation $a_c=\frac{v^2}{R_c}\Rightarrow R_c=\frac{v^2}{a_c}$
When the body is at A,
$R_c=\frac{u^2}{gcos\theta }=\frac{20\times 20}{10\times \frac{4}{5}}=50m$
When the body is at B,
$R_c=\frac{v^2}{g}=\frac{(20\cos 37{)}^2}{10}=25.6m$
As the body travels from A to B,
$R_c=\frac{v^2}{a_c}$, $v$ decreases due to loss of KE and $a_c$ increases from $g\cos \theta$ to $g$. Therefore, the value of $R_c$ should continuously decrease. However, as it moves from B to C, $R_c$ will increase continuously.