At what angle should the ball suspended on the thread be deflected so that when it passes the equilibrium position

At what angle should the ball suspended on the thread be deflected so that when it passes the equilibrium position, its acceleration is 5 m / s2?

Given:
a = 5 m / s square
Decision:
rejecting the ball, we raised it to a height h
At this point, it has a potential energy mgh
Then this energy turns into kinetic. At the lowest point, all potential energy has passed into kinetic energy.
Means mvsquare / 2 = mgh
vsquare / 2 = gh
vsquare = 2gh = 2g (R-Rcos alpha) = 2gR (1-cos alpha)
Acceleration is related to speed by the formula
a = vsquare / R
Hence
v square = aR
Then
2gR (1-cos alpha) = aR
2g (1-cos alpha) = a
1-cosalpha = a / (2g)
cos alpha = 1-a / (2g)
alpha = arccos (1-a / (2g)) = arccos (1-5 / (2 * 9.8)) = 41.8 degrees
Answer: 41.8 degrees.