Daniel B. answered 10/30/20
A retired computer professional to teach math, physics
Let
h = 65.2cm be the length of the rod
m be the mass of the ball
v be the minimum speed to make the ball go over the top
g = 9.8 m/s2 be gravitational acceleration
We use conservation of energy.
The kinetic energy imparted to the ball at the bottom (mv2/2)
needs to convert to the potential energy of the ball at the top (2hmg)
mv2/2 = 2hmg
v = sqrt(4hg) = 2sqrt(hg)
= 2sqrt(65.2 cm . 9.8 m/s2)
= 2sqrt(65.2 . 0.01 m . 9.8 m/s2)
= 5 m/s