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4-1.Newton's Laws of Motion
normal
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
D$300\, m/sec$, in horizontal direction
Solution
By $COLM$ in $Hz$ $dir^n$
$\mathrm{m}(100)=\frac{\mathrm{m}}{3} \mathrm{v}^{\prime}$
$\mathrm{V}^{\prime}=300 \mathrm{m} / \mathrm{sin} \mathrm{Hz} \mathrm{dir}^{\mathrm{n}}$
$\mathrm{m}(100)=\frac{\mathrm{m}}{3} \mathrm{v}^{\prime}$
$\mathrm{V}^{\prime}=300 \mathrm{m} / \mathrm{sin} \mathrm{Hz} \mathrm{dir}^{\mathrm{n}}$
Standard 11
Physics
