A 15 g bullet is fired at 630 m/s into a 4.9 kg block that sits at the edge of a 75-cm-high table. The bullet embeds itself in the block and carries it off the table.

How far from the point directly below the table's edge does the block land?

The horizontal distance traveled below the table's edge is 29.4 cm.

The given parameters;

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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