Ive a problem? It so easy but I can't get it

  • #1
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PLEASE HELP A 68 kg acrobat jumps from a 2 meter ledge on to a trampoline. If the acrobat bounces off the trampoline with a velocity of 10.6 m/s over a contact time of 1.02 seconds, what force did the trampoline exert on the acrobat?


Well I ve tried to do it multiples times, using the formula Vf=Vi+a*t and solving for a. Then I substitute in the Force equation but I get the wrong number and my teacher told me is the wrong equation but i don't know what else to use.
 

Answers and Replies

  • #2
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What you want to look up is the definition of impulse...
 
  • #3
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I'm not sure we've been using this definition. But eventually, how can I use it?
Shouldn't I find force?
 
  • #4
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Sure, if you haven't heard of impulse you can start with Newton's 2nd law dp/dt = F, which can also be more loosely written:
Δp = F Δt
 
  • #5
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So in this case do I use that formula and calculate: Δp= F (found with acceleration) * 1.02s?
So is Δp the force that the problem is asking me?
 
  • #6
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Δp/Δt is the force, yes.
And yes, Δt is the value you are given, but the "tricky" part is determining Δp because you are only given the value of p when the acrobat is bouncing off but have to determine p when he's coming on the trampoline...
 
  • #7
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Is it different because the velocity changes right?
For the first p I got p=68*10.6=720.8. So then how am I supposed to find the second one that you were talking about?
 
  • #8
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Well, what's the acceleration of a falling body near the surface of the earth?
 
  • #9
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9.8 m/s^2 ?
 
  • #10
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correct, so what is v(t) for a falling object?
how long will it take the object to fall 2 meters if it starts from rest?
 
  • #11
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If he jumps off a ledge 2 m above the trampoline, what do you think his velocity was immediately before he made contact with the trampoline?

Chet
 

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