Woman swinging across a ravine on a rope

[donkey11]

Homework Statement

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A 5.0 meter long rope hangs vertically from a tree right at the edge of a ravine. The ravine is 3 meter wide, and a woman wants to use the rope to swing to the other side. She runs as fast as she can, grabs the rope, and swings over the ravine.
a) As she swings what energy conservation takes place?
b) When she is directly over the far edge of the ravine, how much higher is she than when she started?
c)Given your answers in part a and part b, how fast must she be running when she grabs the rope in order to swing all the way across the ravine?

Homework Equations

ke= 1/2mv^2
Ug= mgy

The Attempt at a Solution

a) kinetic to potential
b) 1/2 mv^2= mgy
y= 1/2mv^2/ mg
y= mv/mg
But we don't know V

 

[donkey11]

yeah i know i know them but formula would i use for that

 

[donkey11]

i don't know the original speed that she grabs it

 

[donkey11]

yeah but how do i find theta for that

 

[fliinghier]

you know two parts of the triangle.

try the pythagorean theorum

 

[donkey11]

yeah but is it a triangle ?

 

[donkey11]

wont the 5 meters swinging change the length of the hypoteneuse

 

[fliinghier]

what is the huypotenuse?

 

[donkey11]

would be c of pythagoran theorum hich would be the value of 5.83 meters

 

[fliinghier]

you should look at the picture and apply it to your problem. the point of the triangle is where the lady ends up.

 

[donkey11]

yes i see what you mean by that but what is 2m, is x the 5 meters or is 2m

 

[fliinghier]

2m in your problem is the rope

 

[donkey11]

okay so now I am lost 2m is the long one and x is the shorter distance 3 meters, how do i find height

 

[fliinghier]

no, x is your height. the width of the ravine is perpendicular to the height and the length of the rope when it's hanging straight down.

 

[donkey11]

yes i know so assuming i want to use the diagram you gave to solve for X i would take C^2= A^2 + B^2 where C is going to be height nad A is going to length of rope and B is going to be distance across ravine

 

[fliinghier]

is the rope vertical when you get across?

 

[donkey11]

no i would say it is not because it is attached to one tree, like if i use pythagorean theorum i get a value of 5.83 m so would my height be 0.83 meters

 

[fliinghier]

no. if you look closely, the horizontal line is the distance across the ravine, and the 2m is your rope before you go. what are the other two lines?

hint: the hypotenuse in the picture is 2m

 

[donkey11]

well simplly is the rope is 5 meters then by what you siad 2m is equal to 5 meters... so my height would be square root of rope^2- Distance ^2

 

[fliinghier]

almost. you are calculating the distance vertically from the top of the tree.

 

[donkey11]

where should i be calculating the distance from

 

[donkey11]

like how do you get the value for these