A) 3h
B) \[\infty \]
C) \[\frac{5}{3}h\]
D) \[\frac{8}{3}h\]
Correct Answer: A
Solution :
\[\frac{1}{2}mv{{'}^{2}}=\frac{1}{2}\frac{1}{2}m{{v}^{2}}\] \[v'=\frac{v}{\sqrt{2}}\] \[v=eu\] \[e=\frac{1}{\sqrt{2}}\] \[H=\lambda \left( \frac{1+{{e}^{2}}}{1-{{e}^{2}}} \right)\] \[=h\left( \frac{1+\frac{1}{2}}{1-\frac{1}{2}} \right)=3h\]
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