A) zero
B) \[\frac{m{{v}^{3}}}{4\sqrt{2g}}\]
C) \[\frac{m{{v}^{3}}}{\sqrt{2g}}\]
D) \[\frac{m{{v}^{2}}}{2g}\]
Correct Answer: B
Solution :
Maximum height attained by the projectile, \[H=\frac{{{v}^{2}}{{\sin }^{2}}\theta }{2g}\] Here, \[\theta ={{45}^{o}}\] \[\therefore \] \[H=\frac{{{v}^{2}}{{\sin }^{2}}{{45}^{o}}}{2g}=\frac{{{v}^{2}}}{2g}\times {{\left( \frac{1}{\sqrt{2}} \right)}^{2}}=\frac{{{v}^{2}}}{4g}\] Velocity at maximum height\[=v\cos {{45}^{o}}=\frac{v}{\sqrt{2}}\] Angular momentum at maximum height\[=mvr\] \[=m\times \frac{v}{\sqrt{2}}\times \frac{{{v}^{2}}}{4g}\] \[=\frac{m{{v}^{3}}}{4\sqrt{2}g}\]
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