## 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

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

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

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

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

#### 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?

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

#### 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).

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

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

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

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

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

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

#### 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 $\displaystyle a=50\:m/s^2$

$\displaystyle t_{max}=\frac{v_0}{a}=\frac{26.8\:m/s}{50\:m/s^2}=0.54\:s$
$\displaystyle D_{start}=\frac{\left(v_0\right)^2}{2a}=\frac{\left(26.8\:m/s\right)^2}{2\left(50\:m/s^2\right)}=7.2\:m$
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 $\displaystyle t_{max}$.