A) \[\frac{\sqrt{3}{{u}^{2}}}{2g}\]
B) \[\frac{{{u}^{2}}}{3g}\]
C) \[\frac{3{{u}^{2}}}{2g}\]
D) \[\frac{3{{u}^{2}}}{g}\]
Correct Answer: A
Solution :
There is only horizontal component of velocity at the highest point because at highest point vertical component of velocity is zero, body begins to come downwards from the highest point. \[\therefore \] \[u\,\cos \,\theta =\frac{u}{2}\] \[\,\cos \,\theta =\frac{1}{2}=\cos \,\,{{60}^{o}}\] \[\Rightarrow \] \[\theta ={{60}^{o}}\] Range \[R=\frac{2{{u}^{2}}\,\sin \theta \,cos\theta }{g}\] \[R=\frac{2{{u}^{2}}\,\times \,\sin \,{{60}^{o}}\,\cos {{60}^{o}}}{g}\] \[R=\frac{\sqrt{3}{{u}^{2}}}{2g}\]
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