## Physics (10th Edition)

$H=0.6m$
1) Before the skater leaves the track: Before the skater leaves the track, she goes up a height of $\Delta h=0.4m$. As she does so, her KE decreases while her PE increases by the same amount, following the principle of energy conservation. Therefore, $$\Delta PE = -\Delta KE$$ $$mg\Delta h=-\frac{1}{2}m\Delta v^2$$ $$g\Delta h=-\frac{1}{2}(v_f^2-v_0^2) (1)$$ We know $g=9.8m/s^2$, $\Delta h = 0.4m$ and $v_0=5.4m/s$. We can find $v_f$, the speed she leaves the track: $$-\frac{1}{2}(v_f^2-5.4^2)=3.92$$ $$v_f=4.62m/s$$ 2) After she leaves the track: The vertical speed the skater leaves the track is $v_y=4.62\times\sin48=3.43m/s$ Dealing only with the vertical speed, we know the skater reaches her highest point when $v_y=0$. So we take $v_0=3.43m/s$, $v_f=0$ and $g=9.8m/s^2$; we find H. As the process of going up still continues, we continue to employ equation (1): $$H=\Delta h=\frac{-\frac{1}{2}(v_f^2-v_0^2)}{g}=0.6m$$