P02L2018 – Flying Trapeze
Flying Trapeze
In a trapeze circus act, person A propels herself with an unknown velocity at an angle of $53{}^\circ $ above the horizontal and is supposed to be caught by person B whose hands are $6.1\ m$ above and $8.2\ m$ horizontally from her launch point. You can ignore air resistance and assume person A is a point mass.
1. What initial velocity ${\overrightarrow{v}}_0$ is necessary for person A to just reach the hands of person B?
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Person A just reaches the hands of person B is she reaches her maximum height at the location of the hands of person B. Recalling that $v_y=0$ at the maximum height, we write
\begin{aligned}
v^2_{fy}=v^2_{0y}+2a_y\mathrm{\Delta }y\ \ \ \ \ &\Rightarrow \ \ \ \ \ \ 0=v^2_0{{\mathrm{sin}}^2 \left(\theta \right)\ }-2gh \\
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&\Rightarrow \ \ \ \ \ \ v_0=\sqrt{\frac{2gh}{{{\mathrm{sin}}^{\mathrm{2}} \left(\theta \right)\ }}} \\
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&\Rightarrow \ \ \ \ \ \ v_0=\sqrt{\frac{2\cdot 9.8\cdot 6.1}{{{\mathrm{sin}}^2 \left(53\right)\ }}}\approx 13.7\ m/s
\end{aligned}The initial velocity of person A is therefore equal to
\boxed{{\overrightarrow{v}}_0=v_0{\mathrm{cos} \left(\theta \right)\ }\ \hat{i}+v_0{\mathrm{sin} \left(\theta \right)\ }\ \hat{j}=13.7{\mathrm{cos} \left(53\right)\ }\ \hat{i}+13.7{\mathrm{sin} \left(53\right)\ }\ \hat{j}\approx 8.25\ \hat{i}+10.94\ \hat{j}}2. What are the magnitude and direction of person A’s velocity when she reaches person B’s hands?
View answerHide answerPerson A reaches person B’s hands at her maximum height where only the $x$-component of her velocity is nonzero ($v_y=0$ at the max height). Her velocity in the $x$-direction is constant throughout her flight $\left(a_x=0\right)$ and is therefore equal to $v_{0x}$. Thus, we conclude that the magnitude of her velocity when she reaches person B’s hands is equal to
\boxed{v_x=v_0{\mathrm{cos} \left(\theta \right)\ }\approx 8.25\ \ m/s}and the direction of her velocity is horizontal.
3. How much time does it take for person A to reach person B’s hands?
View answerHide answerIn the time $t_1$ it takes person A to reach person B’s hands, she must travel a horizontal distance $d=8.2\ m$ and $t_1$ must therefore satisfy $x\left(t_1\right)=d$ and we solve for $t_1$ as follows
\begin{aligned}
x\left(t_1\right)=d\ \ \ \ \ &\Rightarrow \ \ \ \ \ \ v_0{\mathrm{cos} \left(\theta \right)\ }t_1=d \\
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&\Rightarrow \ \ \ \ \ \ t_1=\frac{d}{v_0{\mathrm{cos} \left(\theta \right)\ }}
\end{aligned}The time $t_1$ it takes person A to reach person B’s hands is equal to
\boxed{t_1=\frac{d}{v_0{\mathrm{cos} \left(\theta \right)\ }}=\frac{8.2}{13.7{\mathrm{cos} \left(53\right)\ }}\approx 1.0\ s}