P02L2018 – Snowball Rolling Off a Roof
Snowball Rolling Off a Roof
A snowball rolls off a roof as shown in the figure. Ignore air resistance.
1. If it doesn’t strike anything while falling, how far from the edge of the building will it strike the ground?
View answerHide answerAs the snowball rolls off the roof, it is launched into the air at an angle $\theta =40{}^\circ $ below the horizontal as shown below
Thus, the kinematic equations of the snowball as it falls are given by
\left\{ \begin{array}{c}
x\left(t\right)=v_0{\mathrm{cos} \left(\theta \right)\ }t \\
\\
v_x\left(t\right)=v_0{\mathrm{cos} \left(\theta \right)\ } \end{array}
\right.\ \ \ \ \ \ \ \ \ \ and\ \ \ \ \ \ \ \ \ \ \left\{ \begin{array}{c}
\displaystyle{y\left(t\right)=-\frac{1}{2}gt^2-v_0{\mathrm{sin} \left(\theta \right)\ }t+H} \\
\\
v_y\left(t\right)=-gt-v_0{\mathrm{sin} \left(\theta \right)\ } \end{array}
\right.where $H=14\ m$ is the height of the roof relative to the ground. Note that the initial velocity in the $y$-direction is negative because the snowball falls downward initially!
If $t_f$ denotes the time the ball strikes the ground, then we have $y\left(t_f\right)=0$ and we solve for $t_f$ as follows
\begin{aligned}
y\left(t_f\right)=0\ \ \ \ &\Rightarrow \ \ \ \ -\frac{1}{2}gt^2_f-v_0{\mathrm{sin} \left(\theta \right)\ }t_f+H=0 \\
\\
&\Rightarrow \ \ \ \ \ \frac{1}{2}gt^2_f+v_0{\mathrm{sin} \left(\theta \right)\ }t_f-H=0 \\
\\
&\Rightarrow \ \ \ \ \left\{ \begin{array}{c}
\displaystyle{t_f=\frac{-v_0{\mathrm{sin} \left(\theta \right)\ }-\sqrt{v^2_0{{\mathrm{sin}}^2 \left(\theta \right)\ }+2gH}}{g}} \\
\\
\displaystyle{t_f=\frac{-v_0{\mathrm{sin} \left(\theta \right)\ }+\sqrt{v^2_0{{\mathrm{sin}}^2 \left(\theta \right)\ }+2gH}}{g}} \end{array}
\right.
\end{aligned}The first root is negative, and we therefore disregard it. Thus, we conclude that the time it takes the snowball to strike the ground is equal to
t_f=\frac{-v_0{\mathrm{sin} \left(\theta \right)\ }+\sqrt{v^2_0{{\mathrm{sin}}^2 \left(\theta \right)\ }+2gH}}{g}=\frac{-7{\mathrm{sin} \left(40\right)\ }+\sqrt{7^2{{\mathrm{sin}}^2 \left(40\right)\ }+2\cdot 9.8\cdot 1.4}}{9.8}\approx 1.30\ sThe distance $R$ from the building at which the snowball strikes the ground is therefore equal to
\boxed{R=x\left(t_f\right)=v_0{\mathrm{cos} \left(\theta \right)\ }t_f=7{\mathrm{cos} \left(40\right)\ }\cdot 1.30\approx 6.97\ m}2. Sketch $x$, $y$ vs. $t$ and $v_x$, $v_y$ vs. $t$ graphs for the snowball as it falls. You do not need to include values from the problem, but your graphs should be clear enough that we know you understand the general concepts.
View answerHide answerThe four graphs are shown below
3. A person is standing the position shown in the figure and is $1.9\ m$ tall. Will this person be hit by the snowball?
View answerHide answerTo determine if the person gets hit by the snowball, we need to determine the height of the snowball at $x=4\ m$ and compare it to the height of the person.
The snowball is located above $x=4\ m$ at time $t_2$ such that $x\left(t_2\right)=4\ m$ and we solve
\begin{aligned}
x\left(t_2\right)=4\ \ \ \ &\Rightarrow \ \ \ \ \ v_0{\mathrm{cos} \left(\theta \right)\ }t_2=4 \\
\\
&\Rightarrow \ \ \ \ \ t_2=\frac{4}{v_0{\mathrm{cos} \left(\theta \right)\ }} \\
\\
&\Rightarrow \ \ \ \ \ t_2=\frac{4}{7\cdot {\mathrm{cos} \left(40\right)\ }}\approx 0.746\ s
\end{aligned}The height of the ball at $t_2\approx 0.746\ s$ is then equal to
\boxed{y\left(t_2\right)=-\frac{1}{2}gt^2_2-v_0{\mathrm{sin} \left(\theta \right)\ }t_2+H=-\frac{1}{2}\cdot 9.8\cdot {0.746}^2-7{\mathrm{sin} \left(40\right)\ }\cdot 0.746+14\approx 7.91\ m}Thus, we conclude that the snowball is well above the person’s head $\left(7.91>1.9\right)$ and that it does not hit the person.