question_answer

A projectile is projected at an angle \[{{60}^{o}}\] with horizontal with speed 10 m/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

Correct Answer: A

Solution :

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