Swing Energy: Solve for Max Speed of 20kg Child

[JoshMP]

Homework Statement

A 20 kg child is on a swing that hangs from 3.0-m-long chains. What is her maximum speed if she swings out to a 45 degree angle?

Homework Equations

KE= 1/2mv^2
Ug= -mgy

The Attempt at a Solution

I tried to use conservation of energy to solve this problem. Ki + Ugi= Kf + Ugf
Ki=0 (v=0), so Ugi= Kf +Ugf. Then I solved for vf and got 7.67 m/s, which is incorrect. What am I doing wrong? Thanks.

 

[Gear300]

The initial KE isn't 0J...if it was, the child wouldn't be able to achieve a new height with the swing.

 

[raiderIV]

Well i would assume that her velocity is at its maximum when her potential energy is 0, which is at the bottom of the swing. We can figure out her potential energy by using the Conservation of energy and some trig:

She hangs 3m from the top of the swing set, therefore we'll set our y-coord 0 point at the spot where the swing is hanging straight up and down.
Now by using:
3-(3*Cos45) - (finding the adjacent side of the right triangle and subtracting it from 3m to get the height she has risen when she is at 45degrees)
we get the height above the 0 point that the girl is.

using ¹/₂mv² = mgh - (The left side of the equation has 0 for potential energy and the right side has 0 for kinetic energy)
you should be able to figure out what her maximum velocity is.

~John