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A ball is released from certain height which losses $50 \%$ of its kinetic energy on striking the ground, it will attain a height again
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The correct answer is:
$\frac{1}{2}$ th of initial height
Ratio of heights
$\frac{h_1}{h_2}=\frac{100}{100-50}=2$
$h_2=\frac{h_1}{2}=\frac{1}{2}$ of the initial height.
$\frac{h_1}{h_2}=\frac{100}{100-50}=2$
$h_2=\frac{h_1}{2}=\frac{1}{2}$ of the initial height.
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