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?
(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
have no idea how to solve this please help
http://www.jiskha.com/search/index.cgi?query=+The+pole+vault+landing+pad+at+an+Olympic+competition+contains+what+is+essentially+a+bag+of+air+that+compresses+from+its+%22resting%22+height+of+1.1+m+down+to+0.1+m+as+the+vaulter+is+slowed+to+a+stop.+
ye i don't understand ho your explaining with the other guy who posted it. i don't know where you get any of your number from
i don't get this either please help me thanks.
i clicked this link and the numbers i see are different than the question. please help
what is energy?
To find the time interval during which a vaulter slows to a stop, we can use the concept of average acceleration. We know that the vaulter starts from a certain height and comes to rest at a lower height, so we can calculate the average acceleration using the formula:
average acceleration = change in velocity / time interval
(a) For the pole vault landing pad that compresses from 1.1 m to 0.1 m as the vaulter is slowed to a stop:
The change in velocity is the difference in the initial vertical velocity (just before the vaulter touches the pad) and the final vertical velocity (when the vaulter comes to rest).
The initial vertical velocity can be calculated using the equation of motion for vertical motion:
v^2 = u^2 + 2as
Where:
v = final vertical velocity (0 m/s, as the vaulter comes to rest)
u = initial vertical velocity
a = acceleration due to gravity (-9.81 m/s^2, assuming the vaulter experiences free-fall motion)
s = displacement (1.1 m - 0.1 m = 1 m, as the pad compresses)
Plugging in the values, we get:
0^2 = u^2 + 2(-9.81)(1)
0 = u^2 - 19.62
u^2 = 19.62
Taking the square root of both sides, we get:
u ≈ 4.43 m/s (approximately)
Now, we can calculate the average acceleration:
average acceleration = change in velocity / time interval
The change in velocity is the initial vertical velocity, u:
average acceleration = u / time interval
Rearranging the equation, we have:
time interval = u / average acceleration
Substituting the values we found:
time interval = 4.43 m/s / (-9.81 m/s^2)
Calculating the result:
time interval ≈ -0.45 s (approximately)
The negative sign indicates that the vaulter is decelerating.
Therefore, the time interval during which a vaulter slows to a stop is approximately 0.45 seconds.
(b) In this case, the vaulter is brought to rest by a 19.2-cm layer of sawdust that compresses to 4.6 cm when he lands:
Using the same method as before, we can calculate the initial vertical velocity:
v^2 = u^2 + 2as
Where:
v = final vertical velocity (0 m/s)
u = initial vertical velocity
a = acceleration due to gravity (-9.81 m/s^2)
s = displacement (19.2 cm - 4.6 cm = 14.6 cm = 0.146 m)
Plugging in the values, we get:
0^2 = u^2 + 2(-9.81)(0.146)
0 = u^2 - 2(9.81)(0.146)
u^2 = 2(9.81)(0.146)
Taking the square root of both sides, we get:
u ≈ 1.76 m/s (approximately)
Now, we can calculate the time interval:
time interval = u / average acceleration
Substituting the values we found:
time interval = 1.76 m/s / (-9.81 m/s^2)
Calculating the result:
time interval ≈ -0.18 s (approximately)
Therefore, the time interval during which the vaulter slows to a stop is approximately 0.18 seconds.