The pole vault landing pad at an Olympic competition contains what is essentiall

Question

The pole vault landing pad at an Olympic competition contains what is essentially a bag of air that compresses from its "resting" height of 1.1 m down to 0.3 m as the vaulter is slowed to a stop.

(a) What is the time interval during which a vaulter who has just cleared a height of 5.7 m slows to a stop?
? s

(b) What is the time interval if instead the vaulter is brought to rest by a 18.0-cm layer of sawdust that compresses to 5.2 cm when he lands?
  ? ms

(c) Qualitatively discuss the difference in average force the vaulter experiences from the two different landing pads. That is, which landing pad would exert the least force on the vaulter and why?

Explanation / Answer

change in potential energy of valuter at 1.1 m = mg( 5.7- 1.1 )

this is equal to kinetic energy at that point .

so , 0,5mv2= mg( 5.7- 1.1 )

=> v= 9.495 m/s

now, stopping distance = (1.1 m-0.3) =0.8m

so , deacceleration

=> V2= ( 2* a*s)

= >a= 56.35 m/s2

so, time = v=u+at

=> t= 9.495/56.35= 0.168 secs

b)

change in potential energy of valuter at 0.18 m = mg( 5.7- 0.18 )

this is equal to kinetic energy at that point .

so , 0,5mv2= mg( 5.7- 0.18)

=> v= 10.40 m/s

now, stopping distance = (0.18-0.052) =0.128m

so , deacceleration

=> V2= ( 2* a*s)

= >a= 422.5 m/s2

so, time = v=u+at

=> t= 10.40/422.5= 0.0246 secs = 24 millisecs

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