Question

A particle is projected at an angle 60 with horizontal with speed 10m/s.The minimum radius of curvature of the trajectory described by the projectile is

Answer 1

radius of the curvature = half of the range
$range = \frac{ {u}^{2} \sin(2 \alpha ) }{g}$
$range = \frac{3600( \sin(120)) }{10}$
$range = 360 \times \frac{ \sqrt{3} }{2}$
$range = 180 \sqrt{3}$
GLAD YOU LIKE IT
radius of the curvature = 90root(3)