A) \[{{\left( \frac{gh}{{{e}^{2}}} \right)}^{1/2}}\]
B) \[{{\left( \frac{2gh}{e} \right)}^{1/2}}\]
C) \[{{\left( \frac{gh}{e} \right)}^{1/2}}\]
D) \[{{\left( \frac{gh}{2e} \right)}^{1/2}}\]
Correct Answer: A
Solution :
\[{{V}_{0}}=\] initial velocity V, V = velocity of ball before and after collision Given \[V{{'}^{2}}=2gh\] Ball losses 50% of energy \[\frac{(1/2)mV{{'}^{2}}}{(1/2)m{{V}^{2}}}=\frac{1}{2}\] \[{{V}^{2}}=2V{{'}^{2}}=4gh\] So \[{{V}^{2}}=V_{0}^{2}+2gh\] Given \[V_{0}^{2}={{V}^{2}}-2gh=4gh-2gh=2gh\] But \[{{\left( \frac{V'}{V} \right)}^{2}}={{e}^{2}}=\left( \frac{2gh}{4gh} \right)=\frac{1}{2}\] \[V_{0}^{2}=2gh=\left( \frac{gh}{1/2} \right)=\frac{gh}{{{e}^{2}}}\] \[{{V}_{0}}=\left( \frac{gh}{{{e}^{2}}} \right)\]
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