Question 11.193: A telemetry system is used to quantify kinematic values of a...

\begin{aligned}\mathbf{v} & =\left(-105 \cos 25^{\circ}-35 \cos 65^{\circ}\right) \mathbf{i}+\left(35 \sin 65^{\circ}-105 \sin 25^{\circ}\right) \mathbf{j} \\& =(-109.95 \mathrm{\ ft} / \mathrm{s}) \mathbf{i}+(-12.654 \mathrm{\ ft} / \mathrm{s}) \mathbf{j} \\\tan \alpha & =\frac{\nu_{y}}{\nu_{x}}=\frac{-12.654}{-109.95} \quad \alpha=6.565^{\circ} \\\nu & =\sqrt{(105)^{2}+(35)^{2}}=110.68 \mathrm{\ ft} / \mathrm{s} \\&\quad\quad\quad\quad\quad\quad\nu=110.7 \mathrm{\ ft} / \mathrm{s}  ⦫ 6.57^{\circ}\blacktriangleleft\end{aligned}

(b) Acceleration of the skier.

\begin{aligned}\mathbf{a} =a_{r} \mathbf{e}_{r}+a_{\theta} \mathbf{e}_{\theta}=\left(\ddot{r}-r \dot{\theta}^{2}\right) \mathbf{e}_{r}+(r \ddot{\theta}+2 \dot{r} \dot{\theta}) \mathbf{e}_{\theta} \\a_{r} =-10-(500)(0.07)^{2}=-12.45 \mathrm{\ ft} / \mathrm{s}^{2} \\a_{\theta} =(500)(0.06)+(2)(-105)(0.07)=15.30 \mathrm{\ ft} / \mathrm{s}^{2} \\ & \mathbf{a}=\left(-12.45 \mathrm{\ ft} / \mathrm{s}^{2}\right) \mathbf{e}_{r}+\left(15.30 \mathrm{~ft} / \mathrm{s}^{2}\right) \mathbf{e}_{\theta}\blacktriangleleft \\ =\left(-17.750 \mathrm{\ ft} / \mathrm{s}^{2}\right) \mathbf{i}+\left(8.6049 \mathrm{\ ft} / \mathrm{s}^{2}\right) \mathbf{j} \\ \mathbf{a} =(-12.45)\left(\mathbf{i} \cos 25^{\circ}+\mathbf{j} \sin 25^{\circ}\right)+(15.30)\left(-\mathbf{i} \cos 65^{\circ}+\mathbf{j} \sin 65^{\circ}\right) \\ \tan \beta =\frac{a_{y}}{a_{x}}=\frac{8.6049}{-17.750} \quad \beta=-25.9^{\circ} \\a =\sqrt{(12.45)^{2}+(15.30)^{2}}=19.725 \mathrm{\ ft} / \mathrm{s}^{2} \\& a=19.73 \mathrm{\ ft} / \mathrm{s}^{2} \ ⦫25.9^{\circ}\blacktriangleleft\end{aligned}

(c)Distance of the jump d.

Projectile motion. Place the origin of the xy-coordinate system at the end of the ramp with the x-coordinate horizontal and positive to the left and the y-coordinate vertical and positive downward.

Horizontal motion: (Uniform motion)

\begin{aligned}x_{0} & =0 \\\dot{x}_{0} & =109.95 \mathrm{\ ft} / \mathrm{s} \quad(\text { from Part } a) \\x & =x_{0}+\dot{x}_{0} t=109.95 t\end{aligned}

Vertical motion: (Uniformly accelerated motion)

\begin{aligned}y_{0} & =0 \\\dot{y}_{0} & =12.654 \mathrm{\ ft} / \mathrm{s} \quad \quad \text { (from Part } a) \\\ddot{y} & =32.2 \mathrm{\ ft} / \mathrm{s}^{2} \\y & =y_{0}+\dot{y}_{0} t+\frac{1}{2} \ddot{y} t^{2}=12.654 t-16.1 t^{2}\end{aligned}

At the landing point,

\begin{aligned}& x=d \cos 30^{\circ} \quad\quad \text{(1)}\\& y=10+d \sin 30^{\circ} \quad \text { or } \quad y-10=d \sin 30^{\circ}\quad\quad \text{(2)}\end{aligned}

Multiply Eq. (1) by \sin 30^{\circ} and Eq. (2) by \cos 30^{\circ} and subtract

\begin{aligned}& x \sin 30^{\circ}-(y-10) \cos 30^{\circ}=0 \\&(109.95 t) \sin 30^{\circ}-\left(12.654 t+16.1 t^{2}-10\right) \cos 30^{\circ}=0 \\&-13.943 t^{2}+44.016 t+8.6603=0 \\& t=-0.1858 \mathrm{~s} \text { and } 3.3427 \mathrm{~s}\end{aligned}

Reject the negative root.

\begin{aligned}& x=(109.95 \mathrm{\ ft} / \mathrm{s})(3.3427 \mathrm{~s})=367.53 \mathrm{\ ft} \\& d=\frac{x}{\cos 30^{\circ}}\quad\quad\quad\quad d=424\mathrm{\ ft}\blacktriangleleft\end{aligned}