Answer:
(a) The velocity (v) of the ball when reaches its maximum altitude is zero .
v= 0
(b) The acceleration of an object free fall motion is constant and is equal to the acceleration due to gravity,then, At maximum height the acceleration of ball is g =-9,8 m/s².
(c)velocity of the ball when it returns to the ground :
in direction -y

(d)a=g = -9,8 m/s² : acceleration of the ball when it returns to the ground
Explanation:
Ball Kinematics
We apply the free fall formula:
vf²=v₀²+2*a*y Formula (1)
y:displacement in meters (m)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration
Data
v₀ = +5.00 m/s
Problem development
(a) What is its velocity when it reaches its maximum altitude?
in ymax, vf=0 , At maximum height the velocity is zero and the ball falls freely
(b) What is its acceleration at this point?
The acceleration of an object free fall motion is constant and is equal to the acceleration due to gravity,then, At maximum height the acceleration of ball is 9,8 m/s².
g= -9.8 m/s²
c) What is the velocity with which it returns to ground level?
We apply the Formula (1) to calculate the maximum height (h):
Ball movement up
vf²=v₀²+2*a*h
0 = (5)² +2*(-9.8)*h
19.6*h = 25
h= 25/19.6 = 1.275 m
Ball movement down
The distance the ball goes up is equal to the distance it goes down
h= 1.275 m , v₀=0
vf²=v₀²+2*a*h
vf²=0+2*(-9.8)*(-1.275)

in direction -y

(d) What is its acceleration at this point?
a is constant , a= g = -9.8 m/s²