(a) Since the ball is moving under the efferct of gravity, the direction of accelration due to travity is always vetically downwards,
(b) At the highest point, the velocity of the ball becomes zero and acceleration is equat to the accelration due to travity=` 9.8 ms^2` in vertically downward direction. ltBrgt (c ) When the highest point is point is chosen as the location for ` x=0 and ` t=0` and vertically downward direction to be the poitive direction of X-axis and upward direction as negative direction of X-axis.
During upwaard motion, sign of position is begative, sign of velocity is positive and sign of acceleration is also positive .
(d0 Let (t0 be the time takn by the ball to reach the highest point where height from ground be (S),
Taking vertical upward motion of the ball,we have, `u =- 29 .4 ms^(-1), a = 9.8 ms^(-2)` v =0 , S = S, t = ?
As, ` v^2 - u^2 aS :. ` 0- (- 29 .4) ^2 = 2 xx 9.8 xx S` or t= (29.4)/(9.8) = 3 s`
Here- ve sign shows that the distance is covered in upward direction. ltBrgt As, ` u- u + at :. 0=- 29 .4 + 9. 8 xx t, or ` t= (29.4)/(9.8) =3s`
It means time of ascent =3 s`
When an object moves under the effect to gravity alone, the time of ascent is alwaus equal to the time of descent.
Therfore total time after which the ball returns to teh player` s hand ` =3 + 3 =6 s`.