Question

A projectile is projected at an angle 45 degree with speed u radius of curvature of trajectory

Answer 1

A projectile is projected at an angle of 45° with horizontal with speed u.

horizontal range = $\frac{u^2sin2\theta}{g}$

here, $\theta=45^{\circ}$

so, horizontal range = $\frac{u^2sin2(45^{\circ})}{g}$

= $\frac{u^2sin90^{\circ}}{g}$

= $\bf{\frac{u^2}{g}}$

we know, horizontal range is maximum displacement covered by particle in horizontal direction.

so, horizontal range = twice the radius of curvature of trajectory.

or, the radius of curvature = half of horizontal range

= $\boxed{\bf{\frac{u^2}{2g}}}$