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.

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

a high pressure system is a dense air mass that is usually cooler and drier than the surrounding air. a low pressure system is a less dense air mass that is usually wetter and warmer than the surrounding air.

in general, areas that experience high atmospheric pressure also experience fair weather. low pressure systems can cause the formation of clouds and storms. air usually flows from areas of high pressure to areas of low pressure..