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Physics

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A short-putter throws the shot with an initial speed of 16 m/s at a 32 degree angle to the horizontal.

Question: Calculate the horizontal distance traveled by the shot if it leaves the athletes hand at a height of 2.05 m above the ground?

- 45.8m was not the correct answer and I have worked the problem several times over

  • Physics -

    Upwards motion
    v=vₒ-g•t1.
    At the top point
    0= vₒ-g•t1,
    t1= vₒ/g=16/9.8=1.63.
    h=vₒ•t1-g•t1²/2 =16•1.63-9.8•(1.63)²/2 =13,1 m.
    L1= vₒ•cos32•t1=16•0.85•1.63=22,12 m.
    H=h+hₒ=13,1 + 2.05=15.15 m.
    H=g•t2²/2.
    t2=sqrt(2H/g) = sqrt(2•15.15/9.8) =1.76 s.
    L2= vₒ•cos32•t2=16•0.85•1.76=23.94 m.
    L =L1+L2 = 22.12+23.94 = 46.06 m.

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