A body is thrown with the velocity `v_0` at an angle of `theta` to the horizon. Determine `v_0 "in" m s^-1` if the maximum height attained by the body is `5 m` and at the highest point of its trajectory the radius of curvature is `r = 3 m`. Neglect air resistance. `[Use sqrt(80) as 9]`.

Answer

`(9)`

Explanation

Radius of curvature at the highest point
`r = (v)/(g) rArr r = ((v_0 cos theta)^2)/(g)`
`v_0 cos theta = (sqrt(g r))`…(i)
Maximum height, `h = (v_0^2 sin^2 theta)/(g) rArr v_0 sin theta = (sqrt(g h))` …(ii)
By using Eqs (i) and (ii), we get
`v_0^2 sin^2 theta + v_0^2 cos^2 theta = gh + gr = g(h + r)`
`v_0 = sqrt(g(h + r))`
`rArr = sqrt(10(5 + 3)) = 9 m s^-1`.