Answer
$\text{The time it takes for the skier to land is}$ $$t=\frac{x}{v_{x}}=\frac{d \cos \alpha}{v_{0} \cos \theta_{0}}=\frac{(7.117 \mathrm{m}) \cos \left(9.0^{\circ}\right)}{(10 \mathrm{m} / \mathrm{s}) \cos \left(11.3^{\circ}\right)}=0.72 \mathrm{s}$$ $\text{Using Eq. $4-23,$ the $x$ -and $y$ -components of the velocity at landing are}$ $$v_{x}=v_{0} \cos \theta_{0}=(10 \mathrm{m} / \mathrm{s}) \cos \left(11.3^{\circ}\right)=9.81 \mathrm{m} / \mathrm{s}$$ $$v_{y}=v_{0} \sin \theta_{0}-g t=(10 \mathrm{m} / \mathrm{s}) \sin \left(11.3^{\circ}\right)-\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)(0.72 \mathrm{s})=-5.07 \mathrm{m} / \mathrm{s}$$ $\text{Thus, the direction of travel at landing is}$ $$ \theta=\tan ^{-1}\left(\frac{v_{y}}{v_{x}}\right)=\tan ^{-1}\left(\frac{-5.07 \mathrm{m} / \mathrm{s}}{9.81 \mathrm{m} / \mathrm{s}}\right)=-27.3^{\circ} $$ $\text{or $27.3^{\circ}$ below the horizontal. }$
The result implies that the angle between the skier's path and the slope is
$$\phi=27.3^{\circ}-9.0^{\circ}=18.3^{\circ},$$
or approximately $18^{\circ}$ to two significant figures.
Work Step by Step
$\text{The time it takes for the skier to land is}$ $$t=\frac{x}{v_{x}}=\frac{d \cos \alpha}{v_{0} \cos \theta_{0}}=\frac{(7.117 \mathrm{m}) \cos \left(9.0^{\circ}\right)}{(10 \mathrm{m} / \mathrm{s}) \cos \left(11.3^{\circ}\right)}=0.72 \mathrm{s}$$ $\text{Using Eq. $4-23,$ the $x$ -and $y$ -components of the velocity at landing are}$ $$v_{x}=v_{0} \cos \theta_{0}=(10 \mathrm{m} / \mathrm{s}) \cos \left(11.3^{\circ}\right)=9.81 \mathrm{m} / \mathrm{s}$$ $$v_{y}=v_{0} \sin \theta_{0}-g t=(10 \mathrm{m} / \mathrm{s}) \sin \left(11.3^{\circ}\right)-\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)(0.72 \mathrm{s})=-5.07 \mathrm{m} / \mathrm{s}$$ $\text{Thus, the direction of travel at landing is}$ $$ \theta=\tan ^{-1}\left(\frac{v_{y}}{v_{x}}\right)=\tan ^{-1}\left(\frac{-5.07 \mathrm{m} / \mathrm{s}}{9.81 \mathrm{m} / \mathrm{s}}\right)=-27.3^{\circ} $$ $\text{or $27.3^{\circ}$ below the horizontal. }$
The result implies that the angle between the skier's path and the slope is
$$\phi=27.3^{\circ}-9.0^{\circ}=18.3^{\circ},$$
or approximately $18^{\circ}$ to two significant figures.
