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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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