A player throws a ball upwards with an initial speed of 29.4 m/s.
(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).
(a) Irrespective of the direction of the motion of the ball, acceleration due to gravity always acts in the downward direction towards the center of the Earth.
(b) At highest point ball comes to rest. So, velocity of the ball at maximum height is Zero.
(c) During upward motion, Position is negative, Velocity is negative and Acceleration is positive.
During downward motion, Position, velocity and acceleration
all are positive.
(d) Initial velocity of the ball, u = 29.4 m/s
Final velocity of the ball, v = 0 m/s
Acceleration due to gravity, g = 9.8 ms-2
For freely falling body, 3rd equation of motion becomes,
v2 – u2 = -2gs
where,
v = Final velocity
u = Initial velocity
g = Acceleration due to gravity
s = Distance covered
∴ s = 
m = 
m = 44.05 m
and, from 1st equation of motion for freely falling body,
v = u - gt
Where,
v = Final velocity
u = Initial velocity
g = Acceleration due to gravity
t = Time
thus, time of ascent = 
s = 3s
For freely falling body, time of ascent = time decent
∴ Total time = 2t = 6 s
Hence, the total time taken by the ball to reach players hands = 6 s.