Two pole-vaulters just clear the bar at the same height. The first lands at a sp

Question

Two pole-vaulters just clear the bar at the same height. The first lands at a speed of 7.60 m/s, and the second lands at a speed of 7.90 m/s. The first vaulter clears the bar at a speed of 2.00 m/s. Ignore air resistance and friction and determine the speed at which the second vaulter clears the bar.

Explanation / Answer

They both clear same height. So change in P.E. for both is same. Let change in P.E. = x and let second person clears the vault at v m/s But change in P.E. = Change in K.E. For first person: change in P.E. = Change in K.E. =>x = 0.5*m*((7.6^2)-(2^2)).....(1) For second person: change in P.E. = Change in K.E. =>x = 0.5*m*((7.9^2)-(v^2))..(2) equating (1) and (2) we get, 0.5*m*((7.6^2)-(2^2)) = 0.5*m*((7.9^2)-(v^2)) Assuming they are having same mass =>v = 2.080 m/s

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