What Is the Speed of the Clown and Performer When They Reach the Launch Height?

[Sneakatone]

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

 

[Doc Al]

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?

 

[Sneakatone]

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.

 

[SammyS]

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.

 

[Sneakatone]

with that speed I found a height of 12m from using 1/2mv^2=mgh,
what am I suppose to do with that?

 

[SammyS]

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.

 

[Sneakatone]

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)

 

[SammyS]

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.