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March 27, 2017

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A basketball is thrown at 45° to the horizontal. The hoop is located 4.5m away horizontally at a height of 1m above the point of release. What is the required initial speed?

  • physics - ,

    Horizontal
    Vcos45*t=4.5
    Vertical:
    Vsin45*t-4.9t^2=1

    solve for t in the first equation (in terms of V), put that into the second, and solve for V

  • physics - ,

    how do u solve v?

  • physics - ,

    Put t=4.5/Vcos45 into the second equation, you only have V as the unknown. It is a quadratic, use the quadratic equation.

  • physics - ,

    Not good explanation

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