Question

A projectile is projected at an angle $60^{\circ }$ with horizontal with speed $10m/s$. The minimum radius of curvature of trajectory described by die projectile is

• A. 2.25 m
• B. 2 m
• C. 10 m
• D. None of these

Answer 1

Answer: (A)

The radius of curvature is given by
$\rho =\frac{m{v}^3}{|\vec{F}\times \vec{V}|}=\frac{m{v}^3}{Fv\sin \theta }$
$=\frac{m{v}^3}{mg\sin \theta }$
$(\because F=mg)$
or $\rho =\frac{v^2}{g\sin \theta }$
$\therefore {\rho }_{\min }=\frac{v^2}{g(\sin \theta {)}_{\max }}$
$\therefore {\rho }_{\min }=v^2/g$
$(\because {\theta }_{\max }=90^{\circ })\theta =90^{\circ }$
is only possible when the projectile is at the maximum height.
At the maximum height,
$v=u\cos \alpha =10\cos 60^{\circ }=5m/s$
$\therefore {\rho }_{\min }=\frac{5\times 5}{9.8}=2.55m$