91Ó°ÊÓ

Acceleration along x and y axes are

$$\begin{gathered} {\mathrm{a}}_{\mathrm{x}}=4.0\mathrm{m}/{\mathrm{s}}^2 \\ {\mathrm{a}}_{\mathrm{y}}=-2.0\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Initial velocity components along x and y axis are

$$\begin{gathered} {\mathrm{v}}_{0\mathrm{x}}=8.0\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_{0\mathrm{y}}=12\mathrm{m}/\mathrm{s} \end{gathered}$$

This problem is based on kinematic equations that describe the motion of an object with constant acceleration. Here, first kinematic equation for vertical motion to find the time for which${\mathbf{v}}_{\mathbf{max}}\mathbf{=}\mathbf{0}\mathbf{m}\mathbf{/}\mathbf{s}$. Using the value obtained for time in the first kinematic equation along x axis, the velocity of the cart when it reaches its greatest y coordinate can be found.

Formula:

The final velocity in kinematic equation can be written as,

${\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_0+\mathrm{at}$ (i)

Where, $v_0$is the initial velocity

The final velocity along y axis is given by,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{y}}={\mathrm{v}}_{0\mathrm{y}}+\mathrm{at} \\ 0=\left(12\mathrm{m}/\mathrm{s}\right)+\left(-2.0\right)\mathrm{t} \\ \mathrm{t}=6.0\mathrm{s} \end{gathered}$$

The final velocity along x axis is given by

$$\begin{gathered} {\mathrm{v}}_{\mathrm{x}}={\mathrm{v}}_{0\mathrm{x}}+\mathrm{at} \\ {\mathrm{v}}_{\mathrm{x}}=\left(\frac{8.0\mathrm{m}}{\mathrm{s}}\right)+\left(\frac{4.0\mathrm{m}}{{\mathrm{s}}^2}\right)\left(6.0\mathrm{s}\right) \\ {\mathrm{v}}_{\mathrm{x}}=32\mathrm{m}/\mathrm{s} \end{gathered}$$