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 \[{{\omega }_{gravity}}=\Delta K=\frac{1}{2}m{{v}^{2}}-\frac{1}{2}m{{(u\,\cos \,\theta )}^{2}}\] \[=\frac{1}{2}m{{u}^{2}}{{\cos }^{2}}\theta {{\tan }^{2}}\frac{\theta }{2}\]You need to login to perform this action.
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