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?
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?
ms
Get speed when the vaulter hit the cushion
(1/2) m v^2 = m g h
v^2 = 2 g h = 2 * 9.8 * (6.9 - 1.1)
v^2 = 134
v = 10.7 m/s
comes to a stop over one meter (1.1-.1)
approximate by constant deacceleration
0 =Vi + a t
1 = Vi t + (1/2) a t^2
a t = -10.7
a = -10.7/t
1 = 10.7 t + (1/2)(-10.7/t) t^2
1 = 10.7 t - 5.35 t
1 = 5.35 t
t = 1/5.35 second
do the second half the same way
i tried to solve for b but got it wrong. can you just tell me how i can get it started please.
i try to answer the problem i am stuck here is the problem 14 8/12 -10/11
To find the time interval during which the vaulter slows to a stop, we can use the concept of free fall. We know that the vaulter falls from a height of 6.9 m and comes to rest at a height of 0.1 m. We'll use the acceleration due to gravity, which is approximately 9.8 m/s².
(a) Finding the time interval with the air bag:
Using the kinematic equation: vf² = vi² + 2ad, where:
- vf is the final velocity (0 m/s because the vaulter comes to rest),
- vi is the initial velocity (which we'll assume to be 0 m/s),
- a is the acceleration due to gravity (-9.8 m/s²),
- d is the displacement (6.9 m - 0.1 m = 6.8 m),
We can rearrange the equation to solve for time (t): t = √((-2d)/a).
Plugging in the values, we have: t = √((-2 * 6.8)/-9.8) = √(13.6/9.8) ≈ √1.3887 ≈ 1.18 s.
Therefore, the time interval is approximately 1.18 s.
(b) Finding the time interval with the layer of sawdust:
Here, we'll use the same kinematic equation, but with different values for displacement (d) and acceleration (a). The displacement is 19.2 cm - 4.6 cm = 14.6 cm.
We need to convert these values to meters before proceeding:
- Displacement: 14.6 cm ÷ 100 = 0.146 m,
- Resting height: 4.6 cm ÷ 100 = 0.046 m.
Now, using the same kinematic equation, we have:
t = √((-2 * 0.146)/-9.8) = √(0.292/9.8) ≈ √0.0298 ≈ 0.173 s.
Therefore, the time interval is approximately 0.173 seconds (or 173 milliseconds).