A tennis ball with a velocity of 24.4 m/s
is thrown perpendicularly at a wall. After
striking the wall, the ball rebounds in the
opposite direction with a speed of 16 m/s.
If the ball is in contact with the wall for
0.013 s, what is the average acceleration of
the ball while it is in contact with the wall?
Answer in units of m/s
2
24.4+16? If it's the change of velocity, wouldn't it be 24.4-16?
a=-16-24.4/.013
a=-40.4/.013
a=3107.69
To find the average acceleration of the ball while it is in contact with the wall, we can use the formula:
Average acceleration = (change in velocity) / (time taken)
First, we need to calculate the change in velocity. The ball initially has a velocity of 24.4 m/s in one direction and after striking the wall, it rebounds in the opposite direction with a speed of 16 m/s.
The change in velocity = final velocity - initial velocity
= -16 m/s - 24.4 m/s
= -40.4 m/s
Next, we need to calculate the time taken for the ball to be in contact with the wall, which is given as 0.013 s.
Now, we can calculate the average acceleration:
Average acceleration = (change in velocity) / (time taken)
= (-40.4 m/s) / (0.013 s)
≈ -3107.7 m/s²
So, the average acceleration of the ball while it is in contact with the wall is approximately -3107.7 m/s² (since it is negative, indicating a deceleration in this case).