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High School Physics

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The men's world record for the shot put, 23.12m , was set by Randy Barnes of the United States on May 20, 1990.

If the shot was launched from 6.00 f above the ground at an initial angle of 42.0 degrees , what was its initial speed?

Someone plz help =-(

  • High School Physics -

    There are two unknowns: the initial speed V and the flight time T. You need two equations to solve for them both.
    The 6 ft launch height should be comverted to meters (1.83 m)
    The horizontal velocity component is Vcos 42 = 0.7431 V
    From the length of the shotput record,
    0.7431 V T = 23.12 m, so VT = 31.11 m

    The vertical component of the initial speed is Vsin42 = 0.6691V
    The time T required to hit the ground is given by
    0.6691 V*T - 4.9 T^2 + 1.83 = 0

    I recommend that you substitute 31.11 m for VT in the last equation, and solve for T. Then use that T to compute V.

  • High School Physics -

    thank you so much!!! i feel so stupid...but at least you helped. Your awesome thanks for saving my day.

  • High School Physics -

    You are welcome. Don't cry about physics. It can be fun.

    To finish the problem, I got
    21.50 -4.9 T^2 + 1.83 = 0
    4.9 T^2 = 19.67
    T = 2.00 s
    V = 31.11/2 = 15.56 m/s

    Most champion shot putters spin their body while they launch the shot with their arm, to get extra velocity and greater distance

  • High School Physics -

    that answer is completely wrong.

  • High School Physics -

    the answer isn't is just rounding errors that give the false answer.if you use all the decimal points, you should get a final answer of 14.5

  • High School Physics -

    you messed up in the finishing part when 21.5 should be added with 1.83 then moved over

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