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A particle of mass is projected with velocity u at an angle \[\theta \] with horizontal. During the period when the particle descends from highest point to the position where its velocity vector makes an angle  \[\frac{\theta }{2}\]with horizontal, the work done by the gravity force is:

A)  \[\frac{1}{2}m{{u}^{2}}{{\tan }^{2}}\frac{\theta }{2}\]        

B)  \[\frac{1}{2}m{{u}^{2}}{{\tan }^{2}}\theta \]

C)  \[\frac{1}{2}m{{u}^{2}}co{{s}^{2}}\theta {{\tan }^{2}}\frac{\theta }{2}\]

D)  \[\frac{\theta }{2}m{{u}^{2}}{{\cos }^{2}}\frac{\theta }{2}{{\sin }^{2}}\theta \]

Correct Answer: C

Solution :

As horizontal component of velocity does not change.             \[\Rightarrow \]                \[V\,\cos \,\theta /2=u\,\cos \,\theta \]                                 \[V=\frac{u\cos \theta }{\cos \frac{\theta }{2}}\]                 and        \[{{\omega }_{gravity}}=\Delta K=\frac{1}{2}m{{v}^{2}}-\frac{1}{2}m{{(u\,\cos \,\theta )}^{2}}\]                                 \[=\frac{1}{2}m{{u}^{2}}{{\cos }^{2}}\theta {{\tan }^{2}}\frac{\theta }{2}\]