How to find initial velocity

  • Thread starter Sneakatone
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  • #26
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I accidentally divided instead,
my new answer is 15.35 m/s=v2
 
  • #27
Nailed it. Now use that velocity in .5*2m*(v_2)^2 to find the maximum height they reach using 2mgh
 
  • #28
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He grabs the person:
PE = mgh_1
KE = .5 (2m)*(v_2)^2
Solve for v_2
This is the speed of just the clown before he grabs the person

Why is the mass of the clown before he grabs the person 2m? :confused:
 
  • #29
Doc Al
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Why is the mass of the clown before he grabs the person 2m? :confused:
It isn't. That was an error.
 
  • #32
hm i got twice that. Not sure what happened. Anyway, now you can use the potential energy at that height to find the velocity at the original height
 
  • #33
Doc Al
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I accidentally divided instead,
my new answer is 15.35 m/s=v2
That's the correct speed of the clown just as she's about to grab the performer. (When she first reaches y = 4.5 m.) What's their speed after she grabs him?

There's no need to calculate the highest point reached (but you can if you like), since what you want is their speed when they fall back to the starting point (y = 0).
 
  • #34
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1/2mv^2=mgh

v=sqrt(gh2)
v=15.34
is this the right method?
 
  • #35
SammyS
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1/2mv^2=mgh

v=sqrt(gh2)
v=15.34
is this the right method?
No.

What is the speed of the pair right after the clown grabs the performer ?

Then what is the kinetic energy of the pair at that moment ?
 
  • #36
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the speed after is 15.34m/s
how would I find KE without knowing what mass is?
 
  • #37
Doc Al
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1/2mv^2=mgh

v=sqrt(gh2)
v=15.34
is this the right method?
How did you get that speed? What does that speed mean?
 
  • #38
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I got speed from using the height 12m in the equation 1/2mv^2=mgh to get v.

the speed means is that it is the velocity where from the original launch height.
 
  • #39
SammyS
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No.

What is the speed of the pair right after the clown grabs the performer ?

Then what is the kinetic energy of the pair at that moment ?

the speed after is 15.34m/s
how would I find KE without knowing what mass is?

Doc pointed out that 15.34m/s is the speed of the clown just before she grabs the performer.



All you know about mass is that the mass of the clown is the same as the mass of the performer. -- but that's enough.
 
  • #40
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with that speed I found a height of 12m from using 1/2mv^2=mgh,
what am I suppose to do with that?
 
  • #41
SammyS
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with that speed I found a height of 12m from using 1/2mv^2=mgh,
what am I suppose to do with that?
What speed are you referring to?

Please use the "Quote" feature of the Forum, or state the value you are referring to, directly in your post.
 
  • #42
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the speed I am talking about is the one I solved v=15.34 (speed of the clown just as she's about to grab the performer)
 
  • #43
SammyS
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the speed I am talking about is the one I solved v=15.34 (speed of the clown just as she's about to grab the performer)
If the clown had not grabbed the performer, but just continued to rise unimpeded, then yes, the maximum height the clown would have achieved would be 12 meters.

You still haven't computed the speed of the clown - performer combination immediately after the clown grabs the performer.

Here's a hint:

The momentum of the clown plus the performer is the same just before the grabbing and just after the grabbing.
 

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