The horizontal distance to the net: "r=11.9 \\text{ m}."
The horizontal component of velocity:
"v_x=v\\cos\\theta,\\\\\\space\\\\\nt=\\frac r{v_x}=\\frac r{v\\cos\\theta}." Meanwhile, during the same time t, the ball must cover the height of
"h=2.5-0.91=1.19\\text{ m}."The vertical velocity at the net is
"u=v\\sin\\theta+gt."
The time is
"t=\\frac{h}{v_\\text{avg}}=\\frac{2d}{2v\\sin\\theta+gt},\\\\\\space\\\\\nht^2+2tv\\sin\\theta-2h=0,\\\\\\space\\\\\n\\frac{hr^2}{v^2\\cos^2\\theta}+2v\\tan\\theta-2h=0,\\\\\\space\\\\\n\\theta=0.05\u00b0."
Comments
Leave a comment