Answer
$v=(32m/s)\hat{i}$
Work Step by Step
We observe two components of the cart's motion.
Vertically, moving up, it reaches a $y_{max}$, stops (going up) for an instant, and then moves down.
In the meantime, it is accelerating horizontally.
The accelerations are constant in either direction, so the Table 1 formulas apply.
$\left[\begin{array}{ll}
v=v_{0}+at, & (2- 11)\\
x-x_{0}=v_{0}t+21at^{2}, & (2- 15) \\
v^{2}=v_{0}^{2}+2a(x-x_{0}) , & (2- 16) \\
x-x_{0}=\frac{1}{2}(v_{0}+v)t, & (2- 17) \\
x-x_{0}=vt-\frac{1}{2}at^{2} & (2- 18)
\end{array}\right]$
Concentrate first on the vertical component.
At the apex (at the greatest y-coordinate $y_{max}$), the $v_{y}$ component of velocity is zero.
Using Table 2-1, we apply $v=v_{0}+at$ to the y-direction in order to find the time needed to reach the apex.
$v_{y}=v_{y0}+a_{y}t$
$0=(12m/s)+(-2.0m/\mathrm{s}^{2})t$
$t=\displaystyle \frac{12m/s}{2.0m/\mathrm{s}^{2}}=6.0s$
In the x-direction, we also apply $v=v_{0}+at$ , , knowing that
the time passed is $t=6.0s:$
$v_{x}=v_{x0}+a_{y}t=(8.0m/s)+(4.0m/\mathrm{s}^{2})(6.0s)=32m/s$
Thus (at the greatest y-coordinate $y_{max}$),
$v=v_{x}\hat{i}+v_{y}\hat{j}=(32m/s)\hat{i}+0$
$v=(32m/s)\hat{i}$
