P02S2019 – Ball Flying Past a Window
Ball Flying Past a Window
Ted is standing in his living room, looking out the window, when he suddenly sees a ball flying across it (see figure below). The ball entered his field of view at point A, and left his field of view at point B. The length and height of the window are $\ell $ and $h$. The ball was in his field of view for a total of $T\ $seconds. I would recommend you choose your origin to be at point A.
1. What is the velocity ${\overrightarrow{v}}_A=v_{Ax}\ \hat{i}+v_{Ay}\ \hat{j}$ of the ball when it entered his field of view at point A? Express it in terms of $\ell ,\ h,\ g,\ $and $T$. You can express your answer in terms of components or in terms of magnitude and direction.
View answer
Choose the origin at point A, we note that the components of the initial velocity of the ball as it enters Ted’s field of view are $v_{Ax}$ and $v_{Ay}$ and therefore the kinematic equations describing the motion of the ball across the window are
\left\{ \begin{array}{c}
x\left(t\right)=v_{Ax}t \\
\\
\displaystyle{y\left(t\right)=-\frac{1}{2}gt^2-v_{Ay}t} \end{array}
\right.At time $t=T$, the ball is a point B and its position therefore satisfies $\displaystyle{\left\{ \begin{array}{c}
x\left(T\right)=\ell \\
y\left(T\right)=-h \end{array}
\right.}$. This allows us to solve for the components $v_{Ax}$ and $v_{Ay}$ as follows.
\begin{aligned}
\left\{ \begin{array}{c}
x\left(T\right)=\ell \\
\\
y\left(T\right)=-h \end{array}
\right.\ \ \ \ &\Rightarrow \ \ \ \ \ \left\{ \begin{array}{c}
v_{Ax}T=\ell \\
\\
\displaystyle{-\frac{1}{2}gT^2-v_{Ay}T=-h} \end{array}
\right. \\
\\
&\Rightarrow \ \ \ \ \ \left\{ \begin{array}{c}
\displaystyle{v_{Ax}=\frac{\ell }{T}} \\
\\
\displaystyle{v_{Ay}=\frac{\displaystyle{h-\frac{gT^2}{2}}}{T}=\frac{h}{T}-\frac{gT}{2}} \end{array}
\right.
\end{aligned}Thus, the velocity of the ball at point A is equal to
\boxed{{\overrightarrow{v}}_A=\frac{\ell }{T}\ \hat{i}-\left(\frac{h}{T}-\frac{gT}{2}\right)\hat{j}}Note: we initially wrote $-v_{Ay}$ in $y\left(t\right)$ for the $y$-component of ${\overrightarrow{v}}_A$. This is because the $y$-velocity should be negative (downward motion) and therefore by writing $-v_{Ay}$ we show our negative sign. However, we cannot forget to include the negative sign in our final answer for ${\overrightarrow{v}}_A$.
2. Assume the ball was thrown horizontally (so that it had no initial velocity in the $y$-direction) from a nearby building (see figure). What is the distance $D$ from point A from which the ball was thrown? Express your answer in terms of $\ell ,\ h,\ g,\ $and $T$. If you cannot get part 1., you may express your answer in terms of the components of ${\overrightarrow{v}}_A$.
View answer
We let ${\overrightarrow{v}}_0$ denote the initial horizontal velocity of the ball when it is thrown and set our origin at the launch point. Letting $t=0$ at the launch point, we derive the following kinematic equations for its position and velocity as a function of time
\left\{ \begin{array}{c}
x\left(t\right)=v_0t \\
\\
\displaystyle{y\left(t\right)=-\frac{1}{2}gt^2} \end{array}
\right.\ \ \ \ \ \ \ \ \ \ \left\{ \begin{array}{c}
v_x\left(t\right)=v_0 \\
\\
v_y\left(t\right)=-gt \end{array}
\right.We let $t_A$ denote the time at which the ball reaches point A and satisfies $\displaystyle{\left\{ \begin{array}{c}
x\left(t_A\right)=D \\
v_y\left(t_A\right)=-v_{Ay} \end{array}
\right.}$ which allows us to solve for $t_A$ as follows
\begin{aligned}
v_y\left(t_A\right)=-v_{Ay}\ \ \ \ &\Rightarrow \ \ \ \ -gt_A=-\left(\frac{h}{T}-\frac{gT}{2}\right) \\
\\
&\Rightarrow \ \ \ \ \ t_A=\frac{h}{gT}-\frac{T}{2}
\end{aligned}We then recall that the $x$-component of velocity remains constant in projectile motion and therefore $v_0=v_{Ax}$. This lets us solve for $D$ as shown below
\begin{aligned}
x\left(t_A\right)=D\ \ \ \ &\Rightarrow \ \ \ \ \ v_0t_A=D \\
\\
&\Rightarrow \ \ \ \ \ v_{Ax}t_A=D \\
\\
&\Rightarrow \ \ \ \ \ \frac{\ell }{T}\cdot \left(\frac{h}{gT}-\frac{T}{2}\right)=D \\
\\
&\Rightarrow \ \ \ \ \ D=\frac{\ell h}{gT^2}-\frac{\ell }{2}
\end{aligned}Thus, the distance $D$ is equal
\boxed{D=\frac{\ell h}{gT^2}-\frac{\ell }{2}}Note: we could also say that in the time $t_1$ the ball falls a height $H$ from its launch point to point A … but we do not know what $H$ is!