A player throws a ball upwards with an initail speed of 129.4 ms^(-1)` (a) What is the direction of acceleration during the upward motion of the ball ? (b) What are the velocity and accleration of the ball at the ighest point os its motion. ( c) Choose the ` =x 0 aand ` t= 0` be the loction and time at its highest point, vertically downward direction to be the positve direction of ` X-axis and give the signe of position, velocity and accelration of the ball during its upward, and downward motion. (d) To what hight does the ball rise and after how long does the ball returns to the plausr`s hands. (Take g= 9.8 ms^(-23)` and begleact air resistance)

Explanation

(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`.