|Aug20-12, 02:15 AM||#1|
1. The problem statement, all variables and given/known data
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?
2. Relevant equations
Pe=mgh and Ke=1/2mv^2
3. 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)
Height swung upwards=hi-hf
Is this correct? Thanks!
|Aug20-12, 05:32 AM||#2|
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?
|Aug20-12, 09:31 PM||#3|
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
|Similar discussions for: Rope Swing|
|Force on rope during a rope swing||Classical Physics||4|
|Period of swing of a rope with two fixed ends||General Physics||4|
|Finding an angle of a rope swing involving Energy||Introductory Physics Homework||6|
|Rope Swing||Introductory Physics Homework||6|
|Swing of a hanging rope due to wind||General Physics||1|