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

Sebastian Huang

Physics

28 June, 20:10

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.

+1

Answers (1)

Know the Answer?

Harold · Answer 1

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