Question

A projectile is projected at an angle of ${45}^o$ with speed u. The radius of curvature of its trajectory at the maximum height is

• A. $\frac{u^2}{2g}$
• B. $\frac{u^2}{4g}$
• C. $\frac{2u^2}{g}$
• D. $\frac{2u^2}{2g}$

Answer 1

The correct option is A $\frac{u^2}{2g}$

Lets consider, $r=$ radius of the curvature

$m=$ mass of the projectile

$g=$ acceleration due to gravity

From the figure,

Force due to gravity $=$ Centripetal force

$mg=\frac{m{v}_x^2}{r}$

$r=\frac{(ucos{45}^0{)}^2}{g}$

$r=\frac{u^2}{g}\times (\frac{1}{\sqrt{2}}{)}^2$

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