Question

A player throws a ball upwards with an initial speed of 29.4 m s^–1.

1. What is the direction of acceleration during the upward motion of the ball?
2. What are the velocity and acceleration of the ball at the highest point of its motion?
3. 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.
4. 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).

Answer 1

1. Since gravity is at work while the ball is in motion, the direction of gravity-related acceleration is always vertically downward.
2. At maximum height, the velocity of the ball becomes zero. Acceleration due to gravity at a given place is constant and acts on the ball at all points (including the highest point) with a constant value, i.e., 9.8 m/s².
3. If we consider highest point of ball motion as x = 0, t = 0 and vertically downward direction as the positive direction of x-axis, then
   • When moving upward, the position sign is negative, the velocity sign is negative, and the acceleration sign is positive, which means that v < 0 and n > 0.
   • When moving downward, the position sign is positive, the velocity sign is positive, and the acceleration sign is also positive, which means that v > 0 and a > 0.

4. 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 = – 9.8 m/s²

   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)/(2xx(-9.8)) = 44.1 m`

   From first equation of motion, time of ascent (t) is given as:

   v = u + at

   `t = (v-u)/a = (-29.4)/-9.8 = 3 s`

   Time of ascent = Time of descent

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