Question

A player throws a ball upwards with an initial speed of 29.4 ms^-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 ms^-2 and neglect air resistance).

Answer 1

1. The direction of acceleration is vertically downwards (towards the Earth)

2. At the highest point, velocity = 0 ms^-1 acceleration = g = 9.8 ms^-2 vertically downwards.

3. During upward motion: 

• position – positive 
• velocity – negative 
• acceleration – positive

During downward motion: 

• position – positive 
• velocity – positive 
• acceleration – positive

4. v = 0 at the highest point

v₀ = 29.4 ms^-1

g = -9.8 ms^-2

v² = v₀² + 2 g h

0 = 29.4² = 2 × (-9.8) × 4

⇒ h = 44.1m

Height of ascension = 44.1 m

v = v₀ + gt

0 = 29.4 – 9.8 t

t = 3 s

Time of ascension = Time of descension

∴ Time for the ball to return = 6 s