Question 13.201: The 2-lb ball at A is suspended by an inextensible cord and ...

\begin{aligned}& \quad\quad x =x_{B} \quad \text { and } \quad t=t_{B} . \quad \text { From Eq. (5), } \\&\left(t_{B}-t_{2}\right) =\frac{\ell \cos \theta-x_{B}}{\nu_{\theta} \sin \theta} & \text{(6)}\end{aligned}

Vertical motion: 

\begin{aligned}\dot{y} & =\nu_{2} \cos \theta-g\left(t-t_{2}\right) \\y & =y_{2}+\left(\nu_{2} \cos \theta\right)\left(t-t_{2}\right)-\frac{1}{2} g\left(t-t_{2}\right)^{2}\end{aligned}

At Point B,

y_{B}=\ell \sin \theta+\left(\nu_{2} \cos \theta\right)\left(t_{B}-t_{2}\right)-\frac{1}{2} g\left(t_{B}-t_{2}\right)^{2}    (7)

Data: \quad\ell=2 \mathrm{\ ft}, \quad x_{B}=0.3 \mathrm{\ ft}, \quad y_{B}=0.4 \mathrm{\ ft}, \quad g=32.2 \mathrm{\ ft} / \mathrm{s}^{2}

With the numerical data,

Eq. (1) becomes \quad\nu_{2}  =\sqrt{64.4 \sin \theta}       (1)′

Eq. (6) becomes \quad t_{B}-t_{2} =\frac{2 \cos \theta-0.3}{\nu_{2} \sin \theta}       (6)′

Eq. (7) becomes \quad  y_{B}=2 \sin \theta+\left(\nu_{2} \cos \theta\right)\left(t_{B}-t_{2}\right)-16.1\left(t_{B}-t_{2}\right)^{2}       (7)′

Method of solution. From a trial value of \theta, calculate \nu_{2} from Eq. (1)′, t_{B}-t_{2} from Eq. (6)’, and y_{B} from Eq. (7)’. Repeat until y_{B}=0.4 \mathrm{\ ft} as required.

Try \theta=30^{\circ}

\begin{aligned}\nu_{2} & =\sqrt{64.4 \sin 30^{\circ}}=5.6745 \mathrm{\ ft} / \mathrm{s} \\t_{B}-t_{2} & =\frac{2 \cos 30^{\circ}-0.3}{5.6745 \sin 30^{\circ}}=0.50473 \mathrm{~s} \\y_{B} & =2 \sin 30^{\circ}+\left(5.6745 \cos 30^{\circ}\right)(0.50473)-(16.1)(0.50473)^{2} \\& =-0.62116 \mathrm{\ ft}\end{aligned}

Try \theta=45^{\circ}.

\begin{aligned}\nu_{2} & =\sqrt{64.4 \sin 45^{\circ}}=6.7482 \\t_{B}-t_{2} & =\frac{2 \cos 45^{\circ}-0.3}{6.7482 \sin 45^{\circ}}=0.23351 \mathrm{~s} \\y_{B} & =2 \sin 45^{\circ}+\left(6.7482 \cos 45^{\circ}\right)(0.23351)-(16.1)(0.23351)^{2} \\& =1.65060 \mathrm{\ ft}\end{aligned}

Try \theta=37.5^{\circ}.

\begin{aligned}\nu_{2} & =\sqrt{64.4 \sin 37.5^{\circ}}=6.2613 \mathrm{\ ft} / \mathrm{s} \\t_{B}-t_{2} & =\frac{2 \cos 37.5^{\circ}-0.3}{6.2613 \sin 37.5^{\circ}}=0.33757 \mathrm{~s} \\y_{B} & =2 \sin 37.5^{\circ}+\left(6.2613 \cos 37.5^{\circ}\right)(0.33757)-(16.1)(0.33757)^{2} \\& =1.05972 \mathrm{\ ft}\end{aligned}

Let u=\theta-30^{\circ}. The following sets of data points have be determined:

\left(u, y_{B}\right)=\left(0^{\circ},-0.62114 \mathrm{\ ft}\right),\left(7.5^{\circ}, 1.05972 \mathrm{\ ft}\right),\left(15^{\circ}, 1.65060 \mathrm{\ ft}\right)

The quadratic curve fit of this data gives

y_{B}=-0.62114+0.29678 u-0.009688711 u^{2}

Setting \quad y_{B}=0.4 \mathrm{\ ft} gives the quadratic equation

-0.009688711 u^{2}+0.29678 u-1.02114=0

Solving for u, \quad u=3.95^{\circ} \text { and } 26.68^{\circ}

Rejecting the second value gives \quad\theta=30^{\circ}+u=33.95^{\circ}.

Try \theta=33.95^{\circ}.

\begin{aligned}& \nu_{2}=\sqrt{64.4 \sin 33.95^{\circ}}=5.997 \mathrm{\ ft} / \mathrm{s} \\& t_{B}-t_{2}=\frac{2 \cos 33.95^{\circ}-0.3}{5.9971 \sin 33.95^{\circ}}=0.40578 \mathrm{~s} \\& y_{B}=2 \sin 33.95^{\circ}+\left(5.997 \cos 33.95^{\circ}\right)(0.40578)-(16.1)(0.40578)^{2} \\&=0.48462 \mathrm{\ ft}\end{aligned}

The new quadratic curve-fit is based on the data points

\left(u, y_{B}\right)=\left(0^{\circ},-0.62114 \mathrm{\ ft}\right),\left(3.95^{\circ}, 0.48462 \mathrm{\ ft}\right),\left(7.5^{\circ}, 1.05972 \mathrm{\ ft}\right) .

The quadratic curve fit of this data is

y_{B}=-0.62114+0.342053907 u-0.015725232 u^{2}

Setting y_{B}=0.4 \mathrm{\ ft} gives

-0.015725232 u^{2}+0.342053907 u-1.02114=0

Solving for u,

u  =3.572^{\circ} \quad \theta=30^{\circ}+3.572^{\circ}=33.572^{\circ}

Try \theta=33.572^{\circ}.

\begin{aligned}\nu_{2} & =\sqrt{64.4 \sin 33.572^{\circ}}=5.9676 \mathrm{\ ft} / \mathrm{s} \\t_{B}-t_{2} & =\frac{2 \cos 33.572^{\circ}-0.3}{5.9676 \sin 33.572^{\circ}}=0.41406 \mathrm{~s} \\y_{B} & =2 \sin 33.572^{\circ}+\left(5.9676 \cos 33.572^{\circ}\right)(0.41406)-(16.1)(0.41406)^{2} \\& =0.40445 \mathrm{\ ft}\end{aligned}

which is close enough to 0.4 ft.

Substituting \theta=33.572^{\circ} and \nu_{2}=5.9676 \mathrm{\ ft} / \mathrm{s} into Eq. (2) along with other data gives

\begin{array}{r}\nu_{0}^{2}=(5.9676)^{2}+(2)(32.2)(2)\left(1+\sin 33.572^{\circ}\right)=235.64 \mathrm{\ ft}^{2} / \mathrm{s}^{2} \\\mathbf{v}_{0}=15.35 \mathrm{\ ft} / \mathrm{s} \longrightarrow\blacktriangleleft\end{array}