Question #105460

[img]https://upload.cc/i1/2020/03/13/uVG8F5.jpg[/img]

Expert's answer

(i)

"h_{max}=\\frac{v_0^2\\sin^2\\alpha}{2g}=\\frac{30^2\\cdot\\sin^250\u00b0}{2\\cdot9.81}=35.14" "m"

(ii)

"L=l_1+l_2=\\frac{v_0^2\\sin2\\alpha}{g}+v_0\\cos\\alpha\\cdot t="

"=\\frac{v_0^2\\sin2\\alpha}{g}+v_0\\cos\\alpha\\cdot \\frac{\\sqrt{2gh+v_0^2}-v_0}{g}="

"=\\frac{30^2\\sin100\u00b0}{9.81}+30\\cos50\u00b0\\cdot \\frac{\\sqrt{2\\cdot 9.81\\cdot 30+(30\\cdot\\sin50\u00b0)^2}-30\\cdot\\sin50\u00b0}{9.81}="

"=90.35+20.48=110.83" "m"

The car will trapp inside the pit EF safely without hitting the vertical wall GF.

(iii)

"v_x=v_0\\cos50\u00b0=30\\cdot\\cos50\u00b0=19.28 m\/s"

"v_y=\\sqrt{2gh+(v_0\\sin50\u00b0)^2}=\\sqrt{2\\cdot 9.81\\cdot 30+(30\\cdot\\sin50\u00b0)^2}="

"=33.42m\/s"

"\\overrightarrow{v}=19.28\\overrightarrow{i}+33.42\\overrightarrow{j}"

"v=\\sqrt{19.28^2+33.42^2}=38.58m\/s"

"\\tan\\alpha=\\frac{v_x}{v_y}=\\frac{19.28}{33.42}=0.5769 \\to\\alpha\\approx30\u00b0"

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