$A$ bullet of mass $m_1$ is moving with speed $v_0$ and hits a sand bag of mass $m_2$. If the speed of the bullet after passing through the sand bag is $\frac{v_0}{3}$,then the height $h$ up to which the bag rises is (assume,$g=$ acceleration due to gravity).

  • A
    $h=\frac{1}{2 g}\left(\frac{2 m_1 v_0}{3 m_2}\right)^2$
  • B
    $h=\frac{2 m_1 v_0}{3 m_2}$
  • C
    $h=\frac{1}{2 g}$
  • D
    $h=\left(\frac{2 m_1 v_0}{3 m_2}\right)^2$

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