(a)The direction of acceleration during the upward motion of the ball is downward.

(b) Velocity = 0, acceleration = 9.8 m/s2

(c) x > 0 for both up and down motions.

v < 0 for up and v > 0 for down motion.

 And a > 0 throughout the motion

44.1 m, 6 s

Explanation:

(a) Irrespective of the direction of the motion of the ball, acceleration (which is actually acceleration due to gravity) always acts in the downward direction towards the centre of the Earth.

(b) At maximum height, the velocity of the ball becomes zero. Acceleration due to gravity at a given place is constant and acts on the ball at all points (including the highest point) with a constant value i.e.,

(c) During upward motion, the sign of position is positive, the sign of velocity is negative, and the sign of acceleration is positive. During downward motion, the signs of position, velocity, and acceleration are all positive.

(d) The initial velocity of the ball, u = 29.4 m/s

The final velocity of the ball, v = 0 (At maximum height, the velocity of the ball becomes zero)

Acceleration, a = – g = $-9.8<mspace\; width="1em"></mspace>m/s^2$

From the third equation of motion, height (h) can be calculated as:

$v^2-u=2gh$
$h=\frac{v^2-u^2}{2g}$
$\Rightarrow h=\frac{{\left(0\right)}^2-{(29.4)}^2}{2\times (-9.8)}=44.1<mspace\; width="1em"></mspace>m$

From the first equation of motion, time of ascent (t) is given as:

$v=u+at$
$t=\frac{v-u}{a}=\frac{-29.4}{-9.8}=3<mspace\; width="1em"></mspace>s$

Now, Time of ascent = Time of descent

Hence, the total time taken by the ball to return to the player’s hands = 3 + 3 = 6 s.