Respuesta :
To calculate the average acceleration of the ball we use the formula,
[tex]a= \frac{v_{f} -v_{i}}{t}[/tex]
Here, [tex]v_{f}[/tex] is the final velocity of the ball and [tex]v_{i}[/tex] is the initial velocity of the ball t is the time in contact with the wall.
Given [tex]v_{i} = 11.3 m/s[/tex] towards the wall and [tex]v_{f} = 7.2207 m/s[/tex]
away from the wall and [tex]t=0.00803 s[/tex].
Substituting these values in above formula , we get
[tex]a=\frac{(-7.2207 m/s)-11.3 m/s}{0.00803 s}[/tex]
Here[tex]v_{f}[/tex] is negative because ball is moving away from the wall.
[tex]a= - 2306 .4 \ m/s^{2}[/tex]
Therefore, average acceleration of the ball is [tex]- 2306 .4 \ m/s^{2}[/tex] (away from the wall).
Average acceleration is the rate of change of velocity with time
The average acceleration of the ball is approximately 2,306.438 m/s²
The reason the above value is correct is as follows:
The known parameters are;
The direction towards the wall = The positive direction
The speed of the tennis ball as it moves towards the wall, v₁ = 11.3 m/s
The speed of the ball in the opposite direction after rebounding from the wall, v₂ = -7.2207 m/s
The time duration the ball is in contact with the wall, Δt = 0.00803 seconds
The required parameter:
The average acceleration of the ball while in contact with the wall
Solution:
[tex]\mathbf{Average \ acceleration, \ a_{av}} = \dfrac{v_2 - v_1}{t_2 - t_1}} = \mathbf{\dfrac{\Delta v }{\Delta t}}[/tex]
The change in velocity, Δv = v₂ - v₁ = -7.2207 m/s - 11.3 m/s = 18.5207 m/s
The change in time = Time elapsed, Δt = 0.00803 s
Therefore;
[tex]Average \ acceleration, \ a_{av} = \mathbf{\dfrac{\Delta v }{\Delta t}} = \dfrac{18.5207 \ \frac{m}{s} }{0.00803 \ s} \approx 2,306.438 \, \dfrac{m}{s^2}[/tex]
The average acceleration of the ball, [tex]a_{av}[/tex] ≈ 2,306.438 m/s²
Learn more about average acceleration here:
https://brainly.com/question/17699427
