P02L2021 – Lunch Throw
Lunch Throw
Your roommate $A$ leaves for her physics class and runs down the sidewalk at velocity ${\overrightarrow{v}}_A$. You then realize that she forgot her lunch and you get the excellent idea to throw it out the window of your 3${}^{rd}$ floor apartment which is directly above the start of the sidewalk. (You are in another section of the same class and both of you understand kinematics very well. It will be fine!). She has been running down the sidewalk for time $t_A$ at the moment you throw the lunch with a velocity ${\overrightarrow{v}}_L$ horizontally from a height $H$ above the sidewalk. Write all answers in terms of variables given in the problem and any appropriate constants.
1. Find the time it will take for the lunch to travel from your hand to a point just before hitting the sidewalk.
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The lunch is launched horizontally $\left(\theta =0\right)$ and therefore its kinematic equations during its flight are given by
\left\{ \begin{array}{c}
x\left(t\right)=v_Lt \\
\\
v_x\left(t\right)=v_L \end{array}
\right.\ \ \ \ \ \ \ \ \ \ and\ \ \ \ \ \ \ \ \ \ \left\{ \begin{array}{c}
\displaystyle{y\left(t\right)=-\frac{1}{2}gt^2+H} \\
\\
v_y\left(t\right)=-gt \end{array}
\right.The lunch will (barely) hit the sidewalk after being in flight a time $t_f$ such that $y\left(t_f\right)=0$ and we solve for $t_f$ as follows
\begin{aligned}
y\left(t_f\right)=0\ \ \ \ &\Rightarrow \ \ \ \ -\frac{1}{2}gt^2_f+H=0 \\
\\
&\Rightarrow \ \ \ \ \ t_f=\sqrt{\frac{2H}{g}}
\end{aligned}The time it takes the lunch to travel from your hand to a point just before hitting the sidewalk is equal to
\boxed{t_f=\sqrt{\frac{2H}{g}}}2. How far down the sidewalk has she traveled at the point when she catches the lunch just before it hits the sidewalk?
View answerHide answerYour friend runs a constant speed $v_A$ for an amount of time $t_A$ before you throw the sandwich. In that amount of time, she travels a distance $v_At_A$. After the sandwich is thrown, she will continue running at constant speed $v_A$ for an amount of time $t_f$ until she catches the lunch. In that amount of time, she will travel a distance $v_At_f$.
Overall, by the time she catches the lunch, she will have traveled a total distance $d$ equal to
\boxed{d=v_At_A+v_At_f=v_A\left[t_A+\sqrt{\frac{2H}{g}}\right]}3. What was the magnitude of the velocity $v_L$ of the lunch at the time you threw it?
View answerHide answerThe speed $v_L$ of the lunch when it is thrown must be such that in an amount of time $t_f$, the lunch travels a total distance $d$ horizontally. In other words, it must satisfy $x\left(t_f\right)=d$ and we solve for $v_L$ as follows
\begin{aligned}
x\left(t_f\right)=d\ \ \ \ &\Rightarrow \ \ \ \ \ v_Lt_f=v_At_A+v_At_f \\
\\
&\Rightarrow \ \ \ \ \ v_L=v_A\frac{t_A}{t_f}+v_A \\
\\
&\Rightarrow \ \ \ \ \ v_L=v_A\left[1+\frac{t_A}{t_f}\right] \\
\\
&\Rightarrow \ \ \ \ \ \boxed{v_L=v_A\left[1+t_A\sqrt{\frac{g}{2H}}\right]}
\end{aligned}4. What are the components of the velocity of the lunch at the moment it was caught by your roommate?
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When the lunch is caught, it has a velocity ${\overrightarrow{v}}_f$ which has an $x$-component equal to $v_x\left(t_f\right)$ and a $y$-component that is equal to $v_y\left(t_f\right)$. Thus, we have
\boxed{
\begin{aligned}
v_{fx}&=v_x\left(t_f\right)=v_L \\
\\
v_{fy}&=v_y\left(t_f\right)=-gt_f=-g\sqrt{\frac{2H}{g}}=-\sqrt{2gH}
\end{aligned}
}