I have been stuck on this for a while now and cant figure out what it is i am doing wrong HELP!!!
A tennis ball with a velocity of 15.5 m/s is
thrown perpendicularly at a wall. After strik-
ing the wall, the ball rebounds in the opposite
direction with a speed of 9.3 m/s.
If the ball is in contact with the wall for
0.015 s, what is the average acceleration of
the ball while it is in contact with the wall?
aq = (V-Vo)/t = (-9.3-15.5) 0.015=-1660
m/s^2.
NOTE: The final velocity is in the opposite direction. Therefore, it is
NEGATIVE.
A 2nd interpretation:
While the ball is in contact with the wall, its' velocity is o.
a = (0-15.5) / 0.015 = 1033.3 m/s^2.
To find the average acceleration of the ball while it is in contact with the wall, you need to use the equation:
average acceleration = (change in velocity) / (time)
In this case, the change in velocity is the difference between the initial velocity and the final velocity of the ball. However, since the ball reverses its direction after striking the wall, the final velocity is negative. Therefore, the change in velocity is:
change in velocity = final velocity - initial velocity
= (-9.3 m/s) - (15.5 m/s)
= -24.8 m/s
The time in contact with the wall is given as 0.015 seconds. Now, you can calculate the average acceleration:
average acceleration = (change in velocity) / (time)
= (-24.8 m/s) / (0.015 s)
= -1653.33 m/s^2
Therefore, the average acceleration of the ball while it is in contact with the wall is -1653.33 m/s^2.