A pole-vaulter runs along the ground towards the horizontal bar brow at 7.25 m/s

Question

A pole-vaulter runs along the ground towards the horizontal bar brow at 7.25 m/s. By rule, the pole blue, L - 4.7 m) has the same mass as the athlece. While running, the pole i held by the athlete at the same height as the runner's own center of mass, so that both the pole and the runner have an initial position of = 1.2 m above the ground. What is the maximum height H2 that the vaulter can reach when the p le rests vertically on the ground (aB seen 1n the . Final' ima below) ? Hint: as seen in the sketch, the final height of the pole is L/2 H2 Initial Final

Explanation / Answer

The given problem can be solved by energy conservation:

Let the mass of pole and man be m each

Let us consider the datum at the ground

initial potential energy of man+initial potential energy of pole=mg(1.2)+mg(1.2)=2.4mg=23.52m

Initial kinetic energy of man+initial kinetic energy of pole=0.5*m*7.252 + 0.5*m*7.252=52.5625m

Total initial energy of man+pole=23.52m+52.5625m=76.0825m

Given ,final height of pole=L/2=4.7/2=2.35m

Let the maximum height reached by runner be H

Final potential energy of man+final potential energy of pole=mgH + mg(L/2)=9.8mH+9.8*m*2.35=9.8mH+23.03m

From conservation of energy, total energy of man+pole should be conserved

Therefore, 76.0825m = 9.8mH+23.03m

H=5.41 meters

Answer is 5.41 meters

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