forces acting on mass A ...
$\displaystyle T - Mg\sin{\theta} = Ma$
forces acting on mass B ...
$\displaystyle mg - T = ma$
combining equations ...
$\displaystyle mg - Mg\sin{\theta} = Ma + ma$
$\displaystyle \frac{m - M\sin{\theta}}{M + m} \cdot g = a$
this acceleration will be constant as mass A slides up 2.8 m ... use an appropriate kinematics equation to find the velocity of mass A when mass B hits the ground.
for the last 0.2 m, the only force acting on mass A will be the component of weight acting parallel to the incline, so mass A's acceleration will change ...
$\displaystyle g\sin{\theta} = a_2$
use the velocity found when mass B hits the ground, the acceleration $\displaystyle a_2$, and the remaining displacement to find the final velocity of mass A when it hits the pulley.
I suggest that you consider conservation of energy principles. There are two phases to this problem - the first phases is the 4 Kg mass being pulled up the ramp by the force of the 3Kg mass, and then the second phase is after the 3Kg mass hits the ground the tension in the rope becomes zero and the 4Kg mass continues to coast up uphill but slows as it rises. For phase 1 you have $\displaystyle \small\Delta KE + \Delta PE = 0$, where $\displaystyle \small \Delta KE = \frac 1 2 (m_A + M_B) v_1^2 $ and $\displaystyle \small \Delta PE = m_A g (2.8m) \sin 30 = m_B g 2.8m$, so you can solve for $\displaystyle v_1$ . For phase 2: $\displaystyle \small \Delta KE = \frac 1 2 m_A (v_2^2-v_1^2) = m_A g (0.2m \sin 30)$
"m" stands for "mass." You are told that the mass of block A is 4 Kg and the mass of block B is 3 Kg.
The change in kinetic energy of an object is: $\displaystyle \frac 1 2 mv_2^2 - \frac 1 2 mv_1^2$.
for the initial motion 2.8 m up the incline ...
$\displaystyle v_f^2 = v_0^2 + 2a(\Delta x)$
$\displaystyle v_f = \sqrt{2a(\Delta x)} = \sqrt{\frac{2g(2.8)}{7}} = \, 2.8 m/s$
for the final 0.2 m up the incline ...
$\displaystyle v_f^2 = 2.8^2 - g(0.2)$
$\displaystyle v_f = 2.42 \, m/s$