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.

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 0.1 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 part a which was .1869 but still cant figure out part b please help me

Answer 1

(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

Answer 2

i tried what you said for part b and it was marked wrong.please help

Answer 3

To find the time interval for part (b), we need to use the same concept as in part (a) but apply it to the new situation.

In part (a), the landing pad compressed from a resting height of 1.1 m to 0.1 m, so we can calculate the distance the vaulter is stopped during this process:

Stopping distance (d) = Resting height (h) - Compressed height

d = 1.1 m - 0.1 m = 1.0 m

Now, let's calculate the time interval (t) using the equation:

Stopping distance (d) = (1/2) * acceleration (a) * time interval squared (t^2)

Here, d = 1.0 m and a = 9.8 m/s^2 (acceleration due to gravity). We need to solve for t.

1.0 m = (1/2) * 9.8 m/s^2 * t^2

Rearranging the equation:

t^2 = (2 * 1.0 m) / (9.8 m/s^2)

t^2 = 0.2041 s^2

Taking the square root of both sides:

t = √(0.2041 s^2)

t ≈ 0.4522 s

So, the time interval for a vaulter to slow to a stop when brought to rest by the 1.1 m to 0.1 m compression of the landing pad is approximately 0.4522 seconds.

Now, let's move on to part (b), where the vaulter is brought to rest by a 19.2 cm layer of sawdust that compresses to 4.6 cm.

Using the same concept as above, the stopping distance is:

d = 19.2 cm - 4.6 cm = 14.6 cm = 0.146 m

Now, we can use the equation:

0.146 m = (1/2) * 9.8 m/s^2 * t^2

Simplifying the equation:

t^2 = (2 * 0.146 m) / (9.8 m/s^2)

t^2 = 0.0297 s^2

Taking the square root of both sides:

t = √(0.0297 s^2)

t ≈ 0.1723 s

Therefore, the time interval for a vaulter to slow to a stop when brought to rest by the 19.2 cm to 4.6 cm compression of the sawdust layer is approximately 0.1723 seconds or equivalently 172.3 milliseconds (ms).