A player throwsa a ball upwards with an initial speed of `29.4 ms^(-1)`.
(i) What is the direction of acceleration during the upwared motion of the ball?
(ii) What are the velocity and acceleration of the ball at the highest point of its motion?
(iii) Choose the x=0 and t=0 to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of X-axis, and give the signs of positive, velocity and acceleration of the ball during its upward, and downward motion.
(iv) To what height does the ball rise and after how long does the ball return to the player's hand?( Take g`=9.8 ms^(-2)`, and neglect air resistance).

Explanation

(a) As the ball is under the effect of gravity, the direction of acceleration due to gravity during the upward motion of ball is vertically downward.
(b) At the highest point the velocity of ball=0 and acceleration will be equal to acceleration due to gravity and is `9.8m s^(-2)` vertically downward.
(c ) As the downward direction is taken to be positive direction of X-axis, then during upward motion, sign of position and velocity are negative while the sign of acceleration is positive
(d) Let t be the time taken by the ball to reach the highest point of motion=time ofascent For upward motion, `u=-29.4m//s, v.=0`
As `v^(2)-u^(2)=2gh`, where h is the maximum height attained by the ball
`(0)^(2)-(-29.4)^(2)=2(9.8)(-h)`
or `h=(29.4xx29.4)/(2xx9.8)=44.1m`
`v.=v+at`
`0=-29.4+(9.8)t`
or `t=(29.4)/(9.8)=3s`
Time of ascent=Time of descent=3s
Total time `=(3+3)=6s`.