The horizontal distance traveled below the table's edge is 29.4 cm.
The given parameters;
- mass of the bullet, = 15 g = 0.015 kg
- speed of the bullet, = 630 m/s
- mass of the block, = 4.9 kg
The final velocity of the bullet-block system is calculated as follows;
[tex]m_1 u_1 + m_2u_1 = v(m_1 + m_2)\\\\0.015(630) + 4.9(0) = v(0.015 + 4.9)\\\\9.45 = 4.915 v\\\\v = \frac{9.45}{4.915} \\\\v = 1.923 \ m/s[/tex]
The time for the bullet-block system to reach the ground from the table is calculated as follows;
[tex]h = v_0t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\2h = gt^2\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 0.75}{9.8} }\\\\t = 0.153 \ s[/tex]
The horizontal distance traveled below the table's edge is calculated as follows;
[tex]X = vt\\\\X = 1.923 \times 0.153\\\\X = 0.294 \ m\\\\X = 29.4 \ cm[/tex]
Thus, the horizontal distance traveled below the table's edge is 29.4 cm.
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