# A Flower Pot Falling Past a Window| University Physics

## Solution:

Part A

The initial velocity of the pot is zero. Find the velocity $\displaystyle v_b$ of the pot at the bottom of the window. Then using the kinematic equation that relates initial and final velocities, acceleration, and distance traveled, you can solve for the distance $\displaystyle h$.

The average velocity of the flower pot as it passes by the window is $\displaystyle v_{avg}=\frac{L_w}{t}$.

As the pot falls past your window, there will be some instant when the pot’s velocity equals the average velocity $\displaystyle v_{avg}$. Recall that, under constant acceleration, velocity changes linearly with time. This means that the average velocity during a time interval will occur at the middle of that time interval. Meaning, the average velocity happened at time, $\displaystyle \frac{t}{2}$.

Considering the motion at the middle and at the bottom of the window. $\displaystyle g=\frac{v_b-v_{avg}}{\frac{t}{2}}$ $\displaystyle v_b=\frac{gt}{2}+v_{ave}$ $\displaystyle v_b=\frac{gt}{2}+\frac{L_w}{t}$

So, we now know that velocity at the bottom of the window. Consider the motion from the top (point of dropped) to the bottom of the window $\displaystyle \left(v_b\right)^2-\left(v_0\right)^2=2g\left(h-0\right)$ $\displaystyle \left(v_b\right)^2=2gh$ $\displaystyle h=\frac{\left(v_b\right)^2}{2g}$ $\displaystyle h=\frac{\left(\frac{gt}{2}+\frac{L_w}{t}\right)^2}{2g}=\frac{\left(\frac{gt^2+2L_w}{2t}\right)^2}{2g}=\frac{\left(gt^2+2L_w\right)^2}{8gt^2}$

Part B $\displaystyle \left(v_{ground}\right)^2-\left(v_b\right)^2=2gh_b$ $\displaystyle \left(v_{ground}\right)^2=2gh_b+\left(v_b\right)^2$ $\displaystyle\left(v_{ground}\right)=\sqrt{2gh_b+\left(v_b\right)^2}$