When a ball is dropped off a cliff in free fall, it has an acceleration of 9.8 m/s^2. what is its acceleration as it gets closer to the ground

If the ball, the cliff, and the ground are all on the Earth, and everything is bathed in an ocean of air, then the ball's acceleration will decrease as it falls, because of the friction of air resistance.  If it has far enough to fall, it's possible that its acceleration may even become zero, and the ball settle on a constant speed (called "terminal velocity") before it hits the ground.

But until we get to College-level Physics and Engineering, we ALWAYS ignore that stuff, and assume NO AIR RESISTANCE.  The ball is in FREE FALL, and the ONLY force acting on it is the force of gravity.   We also assume that the distance of the fall is small enough so that the value of gravity is constant over the entire fall.

Under those assumptions, there's nothing present to change the acceleration of the falling ball.  It's 9.81 m/s² when it rolls off the edge of the cliff, and it's 9.81 m/s² when it hits the ground.

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft