MCQ
A bomb is projected with $200\,m/s$ at an angle $60^o$ with horizontal. At the highest  point, it explodes into three particles of equal masses. One goes vertically upward with  velocity $100\,m/sec$, second particle goes vertically downward with the same velocity as  the first. Then what is the velocity of the third one-
  • A
    $120\, m/sec$ with $60^o$ angle
  • B
    $200 \,m/sec$ with $30^o$ angle
  • C
    $50\, m/sec$, in horizontal direction
  • $300\, m/sec$, in horizontal direction

Answer

Correct option: D.
$300\, m/sec$, in horizontal direction
d
By $COLM$ in $Hz$ $dir^n$

$\mathrm{m}(100)=\frac{\mathrm{m}}{3} \mathrm{v}^{\prime}$

$\mathrm{V}^{\prime}=300 \mathrm{m} / \mathrm{sin} \mathrm{Hz} \mathrm{dir}^{\mathrm{n}}$

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