Question

a 175-lbf pole vaulter carrying a uniform 16-ft, 10-lbf 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.

Answer 1

42 in / 12 in/ft = 3.5 ft

Total height H = 175 lb* 18 ft + 10 lb * 8 ft = 3330 lb-ft

Beginning = (175 + 10) * 3.5 = 647.5 lb-ft

3330 - 647.5 = 2682.5 lb-ft energy required for jump

1/2 m v^2 = 2682.5 lb-ft

m = (175 + 10) / 32 = 5.78 slug

v = (2 * 2682.5 / 5.78)^1/2 = 30.5 ft/sec

Seems unrealistic since 60 mph = 88 ft/sec

30.5 / 88 * 60 = 20.8 mph speed of vaulter

Note 100 yds = 300 ft

300 ft /10 sec = 30 ft/sec

One has to realize that some of the KE arises from action of vaulters muscles