A player throws a ball upwards with an initial speed of 29.4 m/s. To what height does the ball rise and after how long does the ball return to the player’s hands? (Take g = 9.8 m s–2 and neglect air resistance).
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/s2 From third equation of motion, height (s) can be calculated as: v2−u2=2gs s=(v2−u2)/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.
