Asked by Anonymous
A 50 kg pole vaulter falls from rest from a height of 5 meters onto a foam rubber pad. The pole vaulter comes to rest 0.22 seconds after landing on the pad. Calculate the athlete's velocity just before reaching the pad. Answer in m/s.
I am so confused about which formula to use with this problem. I tried doing it like this, but I don't know if it's right:
final velocity= initial velocity+(gravity)x(time)
final velocity= 0m/s + (-10m/s)(0.22s)
I am so confused about which formula to use with this problem. I tried doing it like this, but I don't know if it's right:
final velocity= initial velocity+(gravity)x(time)
final velocity= 0m/s + (-10m/s)(0.22s)
Answers
Answered by
drwls
This is a trick question.
The time it takes to decelerate to zero depends upon the spring constant of the foam material, and they give you no information about that, so you cannot get the answer by using the decleration time. You have to get of from the height that he falls from, as follows
V = sqrt (2 g H) = 9.9 m/s
The time it takes to decelerate to zero depends upon the spring constant of the foam material, and they give you no information about that, so you cannot get the answer by using the decleration time. You have to get of from the height that he falls from, as follows
V = sqrt (2 g H) = 9.9 m/s
Answered by
Anonymous
V(f)^2=v(i)^2 + 2a(y)
So
V(f)^2=0 + 2(9.8)(5)
V(f)^2=100
V(f)≈10 m/s
So
V(f)^2=0 + 2(9.8)(5)
V(f)^2=100
V(f)≈10 m/s
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