physics
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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 0.1 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 6.9 m slows to a stop?
1 s
(b) What is the time interval if instead the vaulter is brought to rest by a 19.2cm layer of sawdust that compresses to 4.6 cm when he lands?
2 ms
i got part a which was .1869 but still cant figure out part b please help me

(Time interval)*(average velocity during compression) = (compression distance)
(a) For H, use 6.91.1 = 5.8 m and a compression distance of X = 1.0 m
T*(1/2)*sqrt(2gH) = X
T = (2X)/sqrt(2gH)= 0.188 s
In part (b), use the same formula, but with H = 6.9  0.19 m and X = 0.146 m 
i tried what you said for part b and it was marked wrong.please help
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