A tennis ball with a speed of 25.9 m/s is
moving perpendicular to a wall. After striking
the wall, the ball rebounds in the opposite
direction with a speed of 21.3157 m/s.
If the ball is in contact with the wall for
0.0133 s, what is the average acceleration of
the ball while it is in contact with the wall?
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Sra
To find the average acceleration of the ball while it is in contact with the wall, we can use the formula:
Acceleration (a) = Change in velocity (Δv) / Time taken (Δt)
First, we need to calculate the change in velocity of the ball, which is the difference between the initial and final velocities. In this case, the initial velocity is 25.9 m/s (moving perpendicular to the wall) and the final velocity is -21.3157 m/s (rebounding in the opposite direction). Since the final velocity is in the opposite direction, we consider it negative. Therefore:
Δv = final velocity - initial velocity
Δv = -21.3157 m/s - 25.9 m/s
Δv = -47.2157 m/s
Next, we need to calculate the time taken for the ball to be in contact with the wall, which is given as 0.0133 s. Therefore:
Δt = 0.0133 s
Now, we can substitute the values into the formula to find the average acceleration:
Acceleration (a) = Δv / Δt
Acceleration (a) = -47.2157 m/s / 0.0133 s
Acceleration (a) ≈ -3550.37 m/s²
So, the average acceleration of the ball while it is in contact with the wall is approximately -3550.37 m/s².
Divide the velocity change by the time interval. Note that the bounceback velocity is negative.
Acceleration = [25.9 -(-21.3157)]/0.0133
= 3550 m/s^2