Question: A drowsy cat spots a flowerpot that sails first up and then down past an open window. The pot is in view for a total of 0.50 s, and the top to bottom height of the window is 2.00 m. How high above the window top does the flowerpot go?

Let t be the time when a flowerpot passes from across the window, which has length 2.00m So, t=0.25 s (0.25 s is total time, that flowerpot is going upward and downward)

${{\mathrm{V}}_{\mathrm{f}}}^2={{\mathrm{V}}_0}^2+2\mathrm{as}$

The problem deals with the kinematic equation of motion in which the motion of an object is described at constant acceleration. Use kinematic equations to find the height of the flowerpot. Considering the flowerpot would be under the influence of gravity only, we know the acceleration of the flowerpot. Using the time and height of the window, we can find the height to which flowerpot can travel up.

Formula:

$$\begin{gathered} ∆y=V_{0}t+\frac{1}{2}a{t}^2 \\ {V}_0=\frac{2-\left({\displaystyle \frac{1}{2}}\times 9.8\times 0.{25}^2\right)}{0.25} \end{gathered}$$

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

Let $\mathrm{v}$be the velocity of flowerpot when it passes to the top of window.

$\mathrm{\Delta y}={\mathrm{v}}_0\mathrm{t}+\frac{1}{2}{\mathrm{gt}}^2$

${\mathrm{v}}_0=\frac{\mathrm{\Delta y}-\frac{1}{2}{\mathrm{gt}}^2}{\mathrm{t}}$

V₀=6.77 m/s

V₀=6.77 m/s is the velocity of the flowerpot at the bottom of the window while going upward.

Now, we need to find total height attained by the flowerpot before it stops.

By using third kinematic equation we can find its maximum height.

$$\begin{gathered} {\mathrm{v}}_{\mathrm{f}}^2={\mathrm{v}}_0^2+2\mathrm{a\Delta y} \\ 0=6.{77}^2-2(9.8)\mathrm{\Delta y} \\ \mathrm{\Delta y}=2.34\mathrm{m} \end{gathered}$$

This is the height attained by the flowerpot above the bottom of the window. Since the height of the window is 2 m , we have 2.34 m height of the flowerpot above the top of the window.