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 19.2-cm layer of sawdust that compresses to 4.6 cm when he lands?
2 ms
i got a but but follwed theses steps below for b and got .025 and it was marked wrong. please help i don't know what i am doing wrong
(Time interval)*(average velocity during compression) = (compression distance)
(a) For H, use 6.9-1.1 = 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
Whay are you posting my previous answer as a question?
sorry. but i posted under the original question and got not response. but i tried this for part b and i got it wrong
To solve part (b), let's follow the same steps as you did for part (a):
1. Determine the values for H and X:
- H = 6.9 - 0.19 m (since the sawdust layer compresses to 4.6 cm, which is 0.046 m)
- X = 0.146 m (given compression distance)
2. Substitute the values into the formula T * (1/2) * sqrt(2gH) = X:
T * (1/2) * sqrt(2 * 9.8 * H) = X
3. Solve for T:
T = (2X) / sqrt(2 * 9.8 * H)
Now, let's calculate the time interval:
T = (2 * 0.146) / sqrt(2 * 9.8 * (6.9 - 0.19))
≈ 0.025 s
So, the time interval for the vaulter to slow down to a stop when landing on a 19.2-cm layer of sawdust that compresses to 4.6 cm is approximately 0.025 seconds or 25 milliseconds.