Answer
$v_0=27m/s$
Work Step by Step
As the trailer goes upward along the ramp, its KE decreases while PE increases. From the principle of energy conservation, we have $$\Delta KE = -\Delta PE$$ $$\frac{1}{2}m\Delta v^2=-mg\Delta h$$ $$\frac{1}{2} (v_f^2-v_0^2)=-g\Delta h$$
The trailer comes to a stop eventually, so $v_f=0$. To find its initial speed, $v_0$, we need to find how much above the starting point the semitrailer has gone up, or $\Delta h$.
As the figure below illustrates, $$\Delta h=154\sin14=37.26m$$
Therefore, $$\frac{1}{2} (0-v_0^2)=-9.8\times37.26=-365.148$$ $$v_0=27m/s$$
