Question

A child's playground swing is supported by chains that are 4.00 m long. If a child in the swing is 0.63 m above the ground and moving at 5.20 m/s when the chains are vertical, what is the maximum height of the swing ? Assume the masses of the chains are negligible.

Answer 1

$y_i =$ initial height of swing $= 0.63m$

$v_i =$ initial speed of swing $= 5.20m/s$

We have to find the maximum height $y_{max}$ of the swing

Applying the principle of conservation of energy

$\Delta K+ \Delta U = 0$

$(\dfrac{1}{2}mv_f^2-\dfrac{1}{2}mv_i^2)+mg\Delta y = 0$

where $v_f = 0$ at the maximum height of the swing:

$-\dfrac{1}{2}mv_i^2+(mgy_{max}-mgy_i) = 0$

$mgy_{max} = mgy_i+\dfrac{1}{2}mv_i^2$

Hence,

$y_{max} = \dfrac{gy_i+\dfrac{1}{2}v_i^2}{g}$

$y_{max} = \dfrac{(9.8)(0.63)+\dfrac{1}{2}(5.20)}{9.8}$

$y_{max} = 0.89m$