Answer:

A player throws a ball upwards with an initial speed of 29.4ms  

−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  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 players hands ? (Take g=9.8m/s  

2

 and neglect air resistance).

Explanation:

Answer:

Initial velocity of the ball, u = 29.4 m/s

Final velocity of the ball, v = 0 (At maximum height, the velocity of the ball becomes zero)

Acceleration, a =g=9.8m/s

2

From third equation of motion, height (s) can be calculated as:

v

2

−u

2

=2gs

s=(v

2

−u

2

)/2g= ((0)

2

−(29.4)

2

)/2(−9.8)=3s

Time of ascent = Time of descent

Hence, the total time taken by the ball to return to the players hands = 3+3=6 s.