A Flower Pot Falling Past a Window| University Physics

Problem:

As you look out of your dorm window, a flower pot suddenly falls past. The pot is visible for a time t, and the vertical length of your window is L_w. Take down to be the positive direction, so that downward velocities are positive and the acceleration due to gravity is the positive quantity g.

Assume that the flower pot was dropped by someone on the floor above you (rather than thrown downward).

PART A. From what height h above the bottom of your window was the flower pot dropped?

ANSWER: h=\frac{\left(gt^2+2L_w\right)^2}{8gt^2}

The initial velocity of the pot is zero. Find the velocity 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 h.

The average velocity of the flower pot as it passes by the window is 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 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, \frac{t}{2}.

Considering the motion at the middle and at the bottom of the window.

g=\frac{v_b-v_{avg}}{\frac{t}{2}}

v_b=\frac{gt}{2}+v_{ave}

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

\left(v_b\right)^2-\left(v_0\right)^2=2g\left(h-0\right)

\left(v_b\right)^2=2gh

h=\frac{\left(v_b\right)^2}{2g}

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. If the bottom of your window is a height h_b above the ground, what is the velocity v_{ground} of the pot as it hits the ground? You may introduce the new variable v_b, the speed at the bottom of the window, defined by

v_b=\frac{gt}{2}+\frac{L_w}{t}

ANSWER: v_{ground}=\sqrt{2gh_b+\left(v_b\right)^2}

\left(v_{ground}\right)^2-\left(v_b\right)^2=2gh_b

\left(v_{ground}\right)^2=2gh_b+\left(v_b\right)^2

\left(v_{ground}\right)=\sqrt{2gh_b+\left(v_b\right)^2}

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