P02R2015 – Hoop Spacing
Hoop Spacing
A projectile is shot from ground level with an initial velocity $v_0=50\ m/s$ at an angle $\theta =30{}^\circ $. Two hoops with their center a height $H=10\ \ m$ from the ground are spaced in such a way that the ball goes through the center of the first hoop on its way up and through the center of the second hoop on its way down. You may assume $g\approx 10\ m/s^2$ in your calculations.
1. What is the spacing $d$ between the hoops?
View answerHide answerThe ball is hit from an initial position $\left\{ \begin{array}{c}
x_0=0 \
y_0=0 \end{array}
\right.$ and the components of the initial velocity ${\overrightarrow{v}}_0$ and of the acceleration due to gravity are
\left\{ \begin{array}{c}
v_{0x}=v_0{\mathrm{cos} \left(\theta \right)\ } \\
v_{0y}=v_0{\mathrm{sin} \left(\theta \right)\ } \end{array}
\right.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ and\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left\{ \begin{array}{c}
a_x=0 \ \ \ \\
a_y=-g \end{array}
\right.The kinematic equations describing the flight of the ball are therefore given by
\ \left\{ \begin{array}{c}
x\left(t\right)=v_0{\mathrm{cos} \left(\theta \right)\ }t \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\
\\
\displaystyle{y\left(t\right)=-\frac{1}{2}gt^2+v_0{\mathrm{sin} \left(\theta \right)\ }t} \end{array}
\right.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left\{ \begin{array}{c}
v_x\left(t\right)=v_0{\mathrm{cos} \left(\theta \right)\ } \ \ \ \ \ \ \ \ \ \ \ \ \\
\\
v_y\left(t\right)=-gt+v_0{\mathrm{sin} \left(\theta \right)\ } \end{array}
\right.
To derive the spacing $d$ between the hoops, we must first derive the times $t_1$ and $t_2>t_1$ at which the ball goes through the center of each hoop and then compute $d=x\left(t_2\right)-x\left(t_1\right)$. To derive $t_1$ and $t_2$, we solve $y\left(t\right)=H$ as follows
\begin{aligned}
y\left(t\right)=H\ \ \ \ &\Rightarrow \ \ \ \ -\frac{1}{2}gt^2+v_0{\mathrm{sin} \left(\theta \right)\ }t=H \\
\\
&\Rightarrow \ \ \ \ -\frac{1}{2}\cdot 10\cdot t^2+50\cdot {\mathrm{sin} \left(30\right)\ }\cdot t-10=0 \\
\\
&\Rightarrow \ \ \ \ 10\cdot t^2-50\cdot t+20=0 \\
\\
&\Rightarrow \ \ \ \ \ \left\{ \begin{array}{c}
\displaystyle{t_1=\frac{50-\sqrt{{50}^2-4\cdot 10\cdot 20}}{20}\approx 0.438\ s} \\
\\
\displaystyle{t_2=\frac{50+\sqrt{{50}^2-4\cdot 10\cdot 20}}{20}\approx 4.561\ s} \end{array}
\right.
\end{aligned}Given the times $t_1$ and $t_2$ at which the projectile goes through each hoop, we conclude that the distance between the two hoops is given by
\begin{aligned}
d&=x\left(t_2\right)-x\left(t_1\right) \\
\\
&=v_0{\mathrm{cos} \left(\theta \right)\ }t_2-v_0{\mathrm{cos} \left(\theta \right)\ }t_1 \\
\\
&=50{\mathrm{cos} \left(30\right)\ }\cdot 4.561-50{\mathrm{cos} \left(30\right)\ }\cdot 0.438 \\
\\
&\approx 178.53\ m
\end{aligned}Thus, we conclude that the spacing between the hoop is equal to
\boxed{d\approx 178.53\ m}