A) \[\frac{2{{u}^{2}}}{3g}\]
B) \[\frac{{{u}^{2}}}{g}\]
C) \[\frac{\sqrt{3}\,{{u}^{2}}}{2g}\]
D) \[\frac{3{{u}^{2}}}{g}\]
Correct Answer: C
Solution :
At maximum height, we have \[u\cos \theta =\frac{u}{2}\] \[\Rightarrow \] \[\cos \theta =\frac{1}{2}\] \[\Rightarrow \] \[\theta =60{}^\circ \] So, \[R=\frac{{{u}^{2}}\sin (2\times 60{}^\circ )}{g}\] \[=\frac{\sqrt{3}\,{{u}^{2}}}{2g}\]
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