A cannon ball is fired with a velocity 200, m/sec at an angle of 60^o with horizontal. At highest po

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4-1.Newton's Laws of Motion

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A cannon ball is fired with a velocity $200\, m/sec$ at an angle of $60^o$ with the horizontal. At the highest point of its flight it explodes into $3$ equal fragments, one going vertically upwards with a velocity $100\, m/sec$, the second one falling vertically downwards with a velocity $100\, m/sec$. The third fragment will be moving with a velocity

A$100\, m/s$ in the horizontal direction

B$300\, m/s$ in the horizontal direction

C$300\, m/s$ in a direction making an angle of $60^o$ with the horizontal

D$200\, m/s$ in a direction making an angle of $60^o$ with the horizontal

Solution

$3 m\left(200 \cos 60^{\circ}\right) \hat{i}=(m \times 100) \hat{\jmath}+(m \times 100)(-\hat{j})+m \vec{v}$
$\overrightarrow{\mathrm{v}}=(300 \hat{\mathrm{i}}) \mathrm{m} / \mathrm{s}$
$300 \mathrm{m} / \mathrm{s}$ in the horizontal direction.

Standard 11

Physics

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Solution

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