Jane Swings Upward: How High?

[d.tran103]

Homework Statement

Jane, looking for Tarzan, is running at top speed (5.5 m/s) and grabs a vine hanging 4.3 m vertically from a tall tree in the jungle. How high can she swing upward?

Homework Equations

Pe=mgh and Ke=1/2mv^2

The Attempt at a Solution

Can someone tell me if I'm doing this correctly? I have one submission left.
So Pe becomes converted to Ke at the bottom of the swing. And then Ke at the bottom of the swing becomes converted back to Pe. Pe at release is 0 since the vine is hanging vertically. So I have Pe(initial)*0+Ke(initial)=Pe(final) or Ke(initial)=Pe(final)

1/2*m*vi^2=m*g*hf
1/2*vi^2=g*hf
1/2*5.5^2=9.8*hf
hf=1.54337 m.

Height swung upwards=hi-hf
h=4.3-1.54337=2.8 m

Is this correct? Thanks!

 

[PhanthomJay]

Well Jane will swing up to a point 1.5 m vertically above her start point. Or 2.8m below the top of the vine. So how high does she swing up?

 

[d.tran103]

Okay thanks. My first answer was 1.5 however, it wasn't accepted. I'm thinking that the logarithm is using some weird value for g? I'll just ask my professor. Thanks again