## Physics (10th Edition)

$H=6.33m$
1) Find the speed of the person as he hits the water: We know the person hits the water $5m$ from the end of the slide after $0.5s$, so his horizontal speed $v_x=5/0.5=10m/s$. He leaves the slide horizontally, so his initial vertical speed is $v_0=0$. Now we find his vertical speed as he hits the water. We have $v_0=0, t=0.5s, g=9.8m/s^2$, so $$v_y=v_0+gt=4.9m/s$$ Eventually, the person's speed as he reaches the water is $$v_f=\sqrt{v_x^2+v_y^2}=11.14m/s$$ 2) According to the principle of energy conservation, $$E_f-E_0=0$$ $$\frac{1}{2}m(v_f^2-v_0^2)+mg(-H)=0$$ (The person slides down, so $h_f-h_0\lt0$) $$\frac{1}{2}(v_f^2-v_0^2)-gH=0$$ From here, we can find $H$: $$H=\frac{v_f^2-v_0^2}{2g}=\frac{11.14^2-0}{19.6}=6.33m$$