A telemetry system is used to quantify kinematic values of a ski jumper immediately before she leaves the ramp. According to the system r = 500 ft, r = -105 ft/s, r = − 10 ft/s², θ = 25°, θ = 0.07 rad/s, θ = 0.06 rad/s². Determine (a) the velocity of the skier immediately before she leaves the

Question

A telemetry system is used to quantify kinematic values of a ski jumper immediately before she leaves the ramp. According to the system [latex] r=500 \mathrm{\ ft}, \dot { r } =-105 \mathrm{\ ft}/{ s }, \ddot { r } =-10 \mathrm{\ ft}/{ s }^{ 2 }, heta ={ 25 }^{ \circ}[/latex],[latex]\dot { heta } =0.07 \mathrm{\ rad}/{ s },\ddot { heta } =0.06 \mathrm{\ rad}/{ s }^{ 2 }[/latex]. Determine (a) the velocity of the skier immediately before she leaves the jump, (b) the acceleration of the skier at this instant, (c) the distance of the jump d neglecting lift and air resistance.

Accepted Answer

(a) Velocity of the skier. [latex]\quad(r=500 \mathrm{\ ft}, heta=25^{\circ})[/latex]

[latex]\mathbf{v}= u_{r} \mathbf{e}_{r}+ u_{ heta} \mathbf{e}_{ heta}=\dot{r} \mathbf{e}_{r}+r \dot{ heta} \mathbf{e}_{ heta}[/latex]

[latex]=(-105 \mathrm{\ ft} / \mathrm{s}) \mathbf{e}_{r}+(500 \mathrm{\ ft})(0.07 \mathrm{\ rad} / \mathrm{s}) \mathbf{e}_{ heta}[/latex]

[latex]\mathbf{v}=(-105 \mathrm{\ ft} / \mathrm{s}) \mathbf{e}_{r}+(35 \mathrm{\ ft} / \mathrm{s}) \mathbf{e}_{ heta}\blacktriangleleft[/latex]

Direction of velocity:

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

(b) Acceleration of the skier.

[latex]\begin{aligned}\mathbf{a} =a_{r} \mathbf{e}_{r}+a_{ heta} \mathbf{e}_{ heta}=\left(\ddot{r}-r \dot{ heta}^{2} ight) \mathbf{e}_{r}+(r \ddot{ heta}+2 \dot{r} \dot{ heta}) \mathbf{e}_{ heta} \a_{r} =-10-(500)(0.07)^{2}=-12.45 \mathrm{\ ft} / \mathrm{s}^{2} \a_{ heta} =(500)(0.06)+(2)(-105)(0.07)=15.30 \mathrm{\ ft} / \mathrm{s}^{2} \ & \mathbf{a}=\left(-12.45 \mathrm{\ ft} / \mathrm{s}^{2} ight) \mathbf{e}_{r}+\left(15.30 \mathrm{~ft} / \mathrm{s}^{2} ight) \mathbf{e}_{ heta}\blacktriangleleft \ =\left(-17.750 \mathrm{\ ft} / \mathrm{s}^{2} ight) \mathbf{i}+\left(8.6049 \mathrm{\ ft} / \mathrm{s}^{2} ight) \mathbf{j} \ \mathbf{a} =(-12.45)\left(\mathbf{i} \cos 25^{\circ}+\mathbf{j} \sin 25^{\circ} ight)+(15.30)\left(-\mathbf{i} \cos 65^{\circ}+\mathbf{j} \sin 65^{\circ} ight) \ an \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}[/latex]

(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)

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

Vertical motion: (Uniformly accelerated motion)

[latex]\begin{aligned}y_{0} & =0 \\dot{y}_{0} & =12.654 \mathrm{\ ft} / \mathrm{s} \quad \quad ext { (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}[/latex]

At the landing point,

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

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

[latex]\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 ight) \cos 30^{\circ}=0 \&-13.943 t^{2}+44.016 t+8.6603=0 \& t=-0.1858 \mathrm{~s} ext { and } 3.3427 \mathrm{~s}\end{aligned}[/latex]

Reject the negative root.

[latex]\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}[/latex]

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

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