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Q2.6: NCERT Physics - Part 1 MOTION IN A STRAIGHT LINE 2.6 Solution: A player throws a ball upwards with an initial speed of 29.4 m…
⚠️ Common Exam Traps & Conceptual Pitfalls
Students often assume the ball’s acceleration becomes zero at the highest point because its velocity is momentarily zero; however, gravity provides a constant downward acceleration $a = -g$ (or $+g$ if downward is taken as positive) throughout the entire trajectory. Keep the sign convention fixed and remember that a zero velocity does not imply zero acceleration.
A player throws a ball upwards with an initial speed of 29.4 m s�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 m s�2 and neglect air resistance).
Step-by-Step Verified Solution
**Answer:** (a) Downward. (b) Velocity = 0, acceleration = 9.8 m/s_ downward. (c) Position x > 0 (both), velocity v < 0 (upward), v > 0 (downward), acceleration a > 0 (both). (d) Height = 44.1 m, total time = 6.0 s.
**Step-by-Step Explanation:**
(a) Acceleration is due to gravity, directed downward.
(b) At the highest point, velocity is zero; acceleration is still 9.8 m/s_ downward.
(c) With x = 0 at highest point and downward positive: position x > 0 always (ball is below highest point); velocity v < 0 during upward motion (moving against positive direction); v > 0 during downward motion; acceleration a > 0 throughout (downward is positive). [citation:11]
(d) u = 29.4 m/s, v = 0 at highest point. Using v = u - gt: 0 = 29.4 - 9.8t _ t = 3 s.
Height h = u_/2g = (29.4)_/(2 _ 9.8) = 864.36/19.6 = 44.1 m.
Time of ascent = time of descent = 3 s. Total time = 6 s. [citation:11]
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