A Flower Pot Falling Past a Window| University Physics


As you look out of your dorm window, a flower pot suddenly falls past. The pot is visible for a time \displaystyle  t, and the vertical length of your window is \displaystyle 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 \displaystyle g.

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

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

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

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


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}


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To demonstrate the tremendous acceleration of a top fuel Drag Racer| University Physics


To demonstrate the tremendous acceleration of a top fuel drag racer, you attempt to run your car into the back of a dragster that is “burning out” at the red light before the start of a race. (Burning out means spinning the tires at high speed to heat the tread and make the rubber sticky.)

You drive at a constant speed of \displaystyle v_0 toward the stopped dragster, not slowing down in the face of the imminent collision. The dragster driver sees you coming but waits until the last instant to put down the hammer, accelerating from the starting line at constant acceleration, \displaystyle a. Let the time at which the dragster starts to accelerate be \displaystyle t=0.

The figure shows a car moving to the right with speed v 0 toward a stopped dragster, which stands in front of the lights facing to the right.

PART A. What is \displaystyle t_{max}, the longest time after the dragster begins to accelerate that you can possibly run into the back of the dragster if you continue at your initial velocity?


Answer:

\displaystyle t_{max}=\frac{v_0}{a}


PART B. Assuming that the dragster has started at the last instant possible (so your front bumper almost hits the rear of the dragster at \displaystyle t=t_{max}), find your distance from the dragster when he started. If you calculate positions on the way to this solution, choose coordinates so that the position of the drag car is 0 at t=0. Remember that you are solving for a distance (which is a magnitude, and can never be negative), not a position (which can be negative).


Answer:

\displaystyle D_{start}=\frac{\left(v_0\right)^2}{2a}

\displaystyle D_{start}=D_{car}-x_d\left(t_{max}\right)

\displaystyle D_{start}=v_0t_{max}-\frac{1}{2}a\left(t_{max}\right)^2

\displaystyle D_{start}=v_0\left(\frac{v_0}{a}\right)-\frac{1}{2}a\left(\frac{v_0}{a}\right)^2

\displaystyle D_{start}=\frac{\left(v_0\right)^2}{a}-\frac{\left(v_0\right)^2}{2a}

\displaystyle D_{start}=\frac{\left(v_0\right)^2}{2a}


PART C. Find numerical values for \displaystyle t_{max} and \displaystyle D_{start} in seconds and meters for the (reasonable) values \displaystyle v_0=60\:mph (26.8 m/s) and \displaystyle a=50\:m/s^2


Answer:

\displaystyle t_{max}=\frac{v_0}{a}=\frac{26.8\:m/s}{50\:m/s^2}=0.54\:s

\displaystyle D_{start}=\frac{\left(v_0\right)^2}{2a}=\frac{\left(26.8\:m/s\right)^2}{2\left(50\:m/s^2\right)}=7.2\:m

The blue curve shows how the car, initially at \displaystyle x_0, continues at constant velocity (blue) and just barely touches the accelerating drag car (red) at \displaystyle t_{max}.


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