A player throws a ball upwards with an initial speed of 29.4 `ms^(-1)`
(a) What is the direction of acceleration during the upward motion of the ball ?
(b) What are the velocity and acceleration of the ball at the highest point of its motion ?
(c) Choose the x = 0 m and t=0 s 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 position, velocity and acceleration of the ball during its upward and downward motion.
(d) To what height does the ball rise and after how long does the ball return to the player's hands ? (Take g = 9.8 `ms^(-2)` and neglect air resistance).

Explanation

(a) It is due to graviation. Hence the direction is downwards.
(b) Velocity will be zero and acceleration will be equal to g. (Where `g = 9.8 ms ^(-2))`
(c ) Now at maximum height `X =0 and t =0.` Downward direction is taken as positive X-axis. Hence, position in upward direction will be negative, velocity will be negative and acceleration will be position, osition velocity and acceleration will be positive.
`(v lt 0,a gt 0)`
(d) Suppose, time taken to reach maximum height = t
Maximum height achieved by ball = H
Initial velocity `v_(0) = - 29.4 ms ^(-1)`
`a = g = 9.8 ms ^(-2)`

The time taken to reach maximum height is same as the time taken to return back to ground. Hence, time taken by ball to return back to player.s hand `= 3 + 3=6s.`