(a) Since the ball is moving under the effect of gravity, the directionof acceleration due to gravity is always vertically downwards.
(b) When the ball is at the highest point of its motion, its velocitybecomes zero and the acceleration is equal to the acceleration due togravity =9.8 m s–2 in vertically downward direction.
(c) When the highest point is chosen as the location for x=0 and t=0 andvertically downward direction to be the positive direction of x-axis.
During upward motion, sign of position is negative, sign of velocity isnegative and the sign of acceleration is positive i.e.v<0,a>0.
During downward motion, sign of position is positive, sign ofvelocity is positive and the sign of acceleration is also positivei.e.v>0,a>0.
(d) Let t = time taken by the ball to reach the highest point.
H = height of the highest point from the ground.
∴ Initial velocity, u=−29.4 m s−1,a=9.8 m s−2
Final velocity v=0,S=H=?,t=?
Using the relation,v2–u2=2aS, we get
02−(−29.4)2=2×9.8H
or H=

−29.4×29.4

2×9.8

=−44.1 m
where –ve sign shows that the distance is covered in upward direction.Using equation v=u+at, we get
0=−29.4+9.8×t
∴t=

29.4

9.8

=3 s i.e. time of ascent =3 s
Also we know that when the object moves under the effect of gravity alone, the time of ascent is always equal to the time of descent.
∴ Total time after which the ball returns to the player’s hand =2t=2×3=6 s.