In a uniform gravitational field g, which is acting vertically downward, a ball is thrown at an angle with horizontal such that the initial (immediately after projection) radius of curvature is 8 times the minimum radius of curvature. If angle of projection is  n×150, then  the value of n is

At an instant of time  for a projectileRadius of curvature =  square of velocity of projectilecomponent of acceleration perpendicular to velocity of projectileFormula of ROC=v2a⊥Minimum radius of curvature will be at highest point.Therefore,ROCinitial=8 ROChighest point​⇒u2gcosθ=8u2cos2θg​⇒cosθ=12​⇒θ=600 given θ=n ×150⇒600=n ×150⇒n=4