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.
Question
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