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

(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

Answer 1

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

Answer 2

i tried to solve for b but got it wrong. can you just tell me how i can get it started please.

Answer 3

i try to answer the problem i am stuck here is the problem 14 8/12 -10/11