Answer
The block lands on the floor a distance of $0.494~m$ from the table.
Work Step by Step
By conservation of momentum, the final momentum of the block and bullet is equal to the initial momentum of the bullet. We can find the speed $v_f$ of the block and bullet just after the collision:
$m_f~v_f = m_0~v_0$
$v_f = \frac{m_0~v_0}{m_f}$
$v_f = \frac{(0.010~kg)(400.0~m/s)}{0.010~kg+4.0~kg}$
$v_f = 0.9975~m/s$
After the collision, the speed of the block and bullet is $0.9975~m/s$
We can find the time to fall 1.2 meters:
$y = \frac{1}{2}gt^2$
$t = \sqrt{\frac{2y}{g}}$
$t = \sqrt{\frac{(2)(1.2~m)}{9.80~m/s^2}}$
$t = 0.495~s$
We can find the horizontal distance the block moves in this time:
$x = v_x~t = (0.9975~m/s)(0.495~s) = 0.494~m$
The block lands on the floor a distance of $0.494~m$ from the table.
