A ball swings in a vertical circle at the end of a rope 1.50 m long. When the ball is 35.9° past the lowest point on its way up, its total acceleration is (-22.5 i + 20.2 j)m/s^2. Find radial acceleration?

30.24m/s

a = sqrt (vx^2+vy^2)

a=sqrt (-22.5)^2+(20.2)^2
a=30.24m/s^2

29.53 m/s^2

To find the radial acceleration of the ball, we need to break down the total acceleration vector into its radial and tangential components.

The radial acceleration is the component of acceleration that is directed towards the center of the circle. It is responsible for keeping the ball in circular motion.

In this problem, we are given the total acceleration vector as (-22.5 i + 20.2 j) m/s^2. The i-component (-22.5) represents the tangential acceleration, while the j-component (20.2) represents the radial acceleration.

So, the radial acceleration (a_r) is 20.2 m/s^2.