What is the radius of curvature of the parabola traced out by the projectile in the previous problem at a point where the particle velocity makes an angle 2 with the horizontal?

Step-by-step solution

Let v be the velocity at the point where it makes an angle 2 with the horizontal. The horizontal component remains unchanged.So, v_m ( 2 )=u v= u ( 2 ) (i) From figure, m g ( 2 )= m v^2 r r= v^2 g 2 Putting the value of v from Eq. (i), we get r= v^2 ^2 g ^2 2