At what angle `theta_(0)` with the horizontal, should a shell be fired if at the top of its trajectory its path has a radius of curvature equal to twice the maximum height of the trajectory

Answer

`0045`

Explanation

At maximum height `R=(u^(2)cos^(2)theta_(0))/g`
Maximum height of projectile `H=(u^(2)sin^(2)theta_(0))/(2g)`
Given `R= 2H`
`(u^(2) cos^(2)theta)/g=2((u^(2)sin^(2)theta)/(2g))Rightarrow tantheta=1 Rightarrow theta=45^(@)`