Answer
The initial speed of the bullet was 340 m/s.
Work Step by Step
We can use $\mu_k$ to find the magnitude of deceleration after the collision:
$ma = mg~\mu_k$
$a = (9.80~m/s^2)(0.28)$
$a = 2.74~m/s^2$
We can use the acceleration to find the speed $v$ of the block as it started to slide:
$v^2 = 2ax$
$v = \sqrt{2ax} = \sqrt{(2)(2.74~m/s^2)(8.5~m)}$
$v = 6.82~m/s$
We can use conservation of momentum to find the initial speed $v_1$ of the bullet:
$m_1~v_1 = (m_1+m_2)~v$
$v_1 = \frac{(m_1+m_2)(v)}{m_1}$
$v_1 = \frac{(0.028~kg+1.35~kg)(6.82~m/s)}{0.028~kg}$
$v_1 = 340~m/s$
The initial speed of the bullet was 340 m/s.