A particle is projected from ground aiming the maximum horizontal range. Ratio of maximum to minimum radius of curvature of the projectile path is(1)

The path of a projectile is a parabola. The radius of curvature at any point of a parabola is given by R = (1 + y'^2)^(3/2) / |y''|, where y' is the first derivative of y with respect to x, and y'' is the second derivative.

For a projectile, the equation of trajectory is y = xtan(θ) - (gx^2) / (2u^2cos^2(θ)), where θ is the angle of projection, u is the initial velocity, g is the acceleration due to gravity, and x and y are the horizontal and vertical distances respectively.

Differentiating this equation twice with respect to x, we get y' = tan(θ) - gx / (u^2cos^2(θ)) and y'' = -g / (u^2*cos^2(θ)).

Substituting these values in the equation for R, we get R = (1 + (tan(θ) - gx / (u^2cos^2(θ)))^2)^(3/2) / |-g / (u^2*cos^2(θ))|.

The maximum and minimum radii of curvature occur at the maximum and minimum points of the parabola, which are at x = 0 and x = u^2*sin(2θ) / g respectively.

Substituting these values of x in the equation for R, we get R_max = (1 + tan^2(θ))^(3/2) / |g / (u^2cos^2(θ))| and R_min = (1 + (tan(θ) - u^2sin(2θ) / (u^2cos^2(θ)))^2)^(3/2) / |-g / (u^2cos^2(θ))|.

The ratio of maximum to minimum radius of curvature is therefore R_max / R_min = (1 + tan^2(θ))^(3/2) / (1 + (tan(θ) - u^2sin(2θ) / (u^2cos^2(θ)))^2)^(3/2).

For maximum horizontal range, θ = 45°, so tan(θ) = 1 and sin(2θ) = 1. Substituting these values, we get R_max / R_min = (1 + 1)^(3/2) / (1 + (1 - 1))^2)^(3/2) = 2^(3/2) / 1 = 2^(3/2).

Therefore, the ratio of maximum to minimum radius of curvature of the projectile path is 2^(3/2).