A tennis ball with a velocity of +10.7m/s to the right is thrown perpendicularly at a wall. After striking the wall, the ball rebounds in the opposite direction with a velocity of -5.9 m/s (to the left). If the ball is in contact with the wall for 0.009s, what is the average acceleration of the ball while it is in contact with the wall?
(Answer is _m/s^2 to the _{left or right})
To find the average acceleration of the ball while it is in contact with the wall, we can use the formula:
Average acceleration (a) = (Change in velocity)/(Time taken)
Let's break down the given information:
Initial velocity of the ball (v1) = +10.7 m/s (to the right)
Final velocity of the ball (v2) = -5.9 m/s (to the left)
Time taken (t) = 0.009s
To calculate the change in velocity, we have to consider that the initial velocity is positive (+10.7 m/s to the right), and the final velocity is negative (-5.9 m/s to the left). Since the direction has changed, we need to account for the signs.
Change in velocity (Δv) = v2 - v1
= -5.9 m/s - (+10.7 m/s)
= -5.9 m/s - 10.7 m/s
= -16.6 m/s (to the left)
Now, substitute the values into the formula to find the average acceleration:
Average acceleration (a) = (-16.6 m/s) / (0.009s)
≈ -1844.44 m/s^2
Therefore, the average acceleration of the ball while it is in contact with the wall is approximately -1844.44 m/s^2 to the left.