Answer
The child's maximum speed 4.2 m/s
Work Step by Step
Let $L$ be the length of the chains. We can find the height $h$ (above the swing's lowest point) when the chains are at a $45^{\circ}$ angle.
$\frac{L-h}{L} = cos(\theta)$
$h = L~[1-cos(\theta)]$
$h = (3.0~m)~[1-cos(45^{\circ})]$
$h = 0.88~m$
The maximum speed will occur at the lowest point when kinetic energy is at a maximum. We can use conservation of energy to find the child's speed at the lowest point. The kinetic energy at the lowest point will be equal to the potential energy at height $h$.
$KE = PE$
$\frac{1}{2}mv^2 = mgh$
$v^2 = 2gh$
$v = \sqrt{2gh}$
$v = \sqrt{(2)(9.80~m/s^2)(0.88~m)}$
$v = 4.2~m/s$
The child's maximum speed 4.2 m/s
