3/129 A 175‐lb pole vaulter carrying a uniform 16‐ft, 10‐lb pole approaches the jump with a velocity v and manages to barely clear the bar s

Question

3/129 A 175‐lb pole vaulter carrying a uniform 16‐ft, 10‐lb pole approaches the jump with a velocity v and manages to barely clear the bar set at a height of 18 ft. As he clears the bar, his velocity and that of the pole are essentially zero. Calculate the minimum possible value of v required for him to make the jump. Both the horizontal pole and the center of gravity of the vaulter are 42 in. above the ground during the approach.

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Thu Thảo 4 years 2021-08-10T10:08:23+00:00 1 Answers 47 views 0

Answers ( )

  1. bonexptip
    0
    2021-08-10T10:10:07+00:00
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    Answer:

    30ft/s

    Explanation:

    U’1-2 = 0

    Using the equation

    T1 + Vg1 = T2 + Vg2

    Take dactum Vg1 = 0 at ground level

    T1 =[ 1/2 × (175 + 10)/32.2] × V^2

    T1 = 2.87V^2

    T2 = 0

    Vg1 = (175 + 10) ×(42/12) = 648ft.lb

    Vg2 = 175(18) + 10(8) = 3230ft.lb

    Therefore, 2.87V^2 + 648 = 0 + 3230

    3230 – 648 = 2.87V^2

    2582 = 2.87V^2

    V^2 = 2582/2.87

    V ^2 = 889.65

    V = Sqrt (889.65)

    V = 29.99 = approximately 30ft/s

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