A simple pendulum consists of a ball at the end of a light string. If the length of the string is 1.4 m and the ball is released from rest at a point where the string makes an angle of 20 degrees with the vertical, what is the maximum speed of the pendulum?

Use the master energy equation:

U_i + K_i + W_nc = U_f + K_f

i represents the initial position, where the mass is released from. f is the final position, at the bottom of the path. Set the zero level for potential energy to be the bottom of the path.

The initial kinetic energy K_i = 0 because the ball is released from rest
The work done by non-conservative forces W_nc = 0 because we neglect air resistance.
The final potential energy U_f = 0 because we set the zero for potential energy to be the bottom of the path. So:

U_i = K_f

mgh = ½ mv_f², where h is the height from which the ball is released.

The m's cancel, leaving:

v_f = (2gh)^½

h can be determined from geometry. If L is the length of the string:

h = L - Lcos(θ) = L [1-cos(θ)]

With L = 1.40 m and θ = 20 degrees, h = 0.0844 m.

This gives v_f = (2gh)^½ = 1.29 m/s