A 55-kg soccer player jumps vertically upwards and heads the 0.45-kg ball as it is descending vertically with a speed of 29 m/s.

If the ball is in contact with the player's head for 16 ms, what is the average acceleration of the ball? (Note that the force of gravity may be ignored during the brief collision time.)

I will assume he just stops the descent of the ball

change of ball velocity = 29 m/s
a = change of velocity/time
= 29/(16*10^-3) = 1.81 * 10^3 m/s^2

To calculate the average acceleration of the ball, we can use the following formula:

average acceleration = change in velocity / time

To find the change in velocity, we need to determine the initial and final velocities of the ball.

Since the ball is descending vertically with a speed of 29 m/s, the initial velocity (vi) is -29 m/s (negative since it's downward).

During the collision, the player's head brings the ball to rest momentarily. Therefore, the final velocity (vf) of the ball is 0 m/s.

Now, we can use the formula for average acceleration:

average acceleration = (vf - vi) / time

Since the final velocity is 0 m/s and the initial velocity is -29 m/s, the change in velocity (vf - vi) is:

0 m/s - (-29 m/s) = 29 m/s

The time given in the problem is 16 ms, which we need to convert to seconds:

16 ms = 16/1000 s = 0.016 s

Finally, we can substitute the values into the formula to calculate the average acceleration:

average acceleration = (29 m/s) / (0.016 s) = 1812.5 m/s²

Therefore, the average acceleration of the ball is 1812.5 m/s².