A 45.5-g Super Ball traveling at 29.0 m/s bounces off a brick wall and rebounds at 19.5 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 4.15 ms, what is the magnitude of the average acceleration of the ball during this time interval?

the super ball will rebound at 0.12745 seconds at a velocity of giga 10 milliseconds.

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V1=29 , v2= -19.5 t = 4.15*10^-3

average a = (29+19.5)/4.15*10^-3

To find the magnitude of the average acceleration of the ball during the time it is in contact with the wall, we can use the equation:

Acceleration = (Final Velocity - Initial Velocity) / Time

Given:
Mass of the ball (m) = 45.5 g = 0.0455 kg
Initial velocity (v1) = 29.0 m/s
Final velocity (v2) = 19.5 m/s
Time (t) = 4.15 ms = 0.00415 s

First, let's convert the mass of the ball from grams to kilograms:
Mass (m) = 0.0455 kg

Next, we can calculate the acceleration using the formula mentioned above:

Acceleration = (Final Velocity - Initial Velocity) / Time
Acceleration = (19.5 m/s - 29.0 m/s) / 0.00415 s

Now, subtract the initial velocity from the final velocity:
Acceleration = (-9.5 m/s) / 0.00415 s

Finally, divide the change in velocity by the time interval to get the average acceleration:
Acceleration ≈ -2298.80 m/s²

The negative sign indicates that the acceleration is in the opposite direction to the initial velocity. To find the magnitude (without the negative sign), we can take the absolute value of the result:

Magnitude of Average Acceleration ≈ abs(-2298.80 m/s²) ≈ 2298.80 m/s²

Therefore, the magnitude of the average acceleration of the ball during the 4.15 ms contact with the wall is approximately 2298.80 m/s².