Question

A player throws a ball upwards with an initial speed of 29.4 m \(\text s^{-1}\). 

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 \(\text m \;\)\(\text s^{-2}\) and neglect air resistance).

Answer 1

Given:

Initial speed of the ball, u = 29.4 m s⁻¹
Acceleration due to gravity, g = 9.8 m s⁻²
Air resistance is neglected.

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(a) Direction of acceleration during upward motion

During the upward motion of the ball, the acceleration is due to gravity.

Gravity always acts vertically downward, irrespective of the direction of motion.

Therefore, the direction of acceleration during upward motion is downward.

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(b) Velocity and acceleration at the highest point

At the highest point of motion:

Velocity of the ball becomes zero.

However, the acceleration due to gravity is still acting.

Acceleration at the highest point:

a = g = 9.8 m s⁻² downward

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(c) Signs of position, velocity, and acceleration

Given:

Origin (x = 0) is chosen at the highest point.
Positive x-axis is vertically downward.

During upward motion:

• Position: Ball is below the origin → x positive
• Velocity: Ball is moving upward → opposite to positive direction → velocity negative
• Acceleration: Gravity acts downward → along positive x-axis → acceleration positive

At the highest point:

• Position: x = 0
• Velocity: v = 0
• Acceleration: positive (g)

During downward motion:

• Position: Ball is below the highest point → x positive
• Velocity: Ball is moving downward → along positive x-axis → velocity positive
• Acceleration: Gravity acts downward → acceleration positive

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(d) Maximum height and time of return

Maximum height:

Using the equation:

v² = u² − 2gh

At the highest point, v = 0

0 = (29.4)² − 2 × 9.8 × h

h = (29.4)² / (2 × 9.8)

h = 44.1 m

Time to return to the player’s hands:

Time taken to reach the highest point:

v = u − gt

0 = 29.4 − 9.8t

t = 3 s

Total time of flight = 2 × 3 = 6 s

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Final Answers:

(a) Acceleration is vertically downward.
(b) Velocity = 0; Acceleration = 9.8 m s⁻² downward.
(c) Signs explained above for upward and downward motion.
(d) Maximum height = 44.1 m; Time of return = 6 s.