A typical baseball has mass of 150 grams. Suppose a batter hits a fastball pitched to him at 92 mph (which is about 40.89 m/s). After contact, the ball's speed is measured as 104 mph (which is about 46.22 m/s) directly back at the pitcher (who is able to duck out of the way, barely). The length of time that the ball is in contact with the bat is 0.002 s. During this time, the only force that is important for the ball's motion is the normal force applied by the bat. Calculate both the ball's acceleration and the force applied on it during this time. (We'll approximate the acceleration and force as constant, which is probably not quite true, but close enough for this problem.)

a=(V-Vo)/t=(46.22-40.89)/0.002=2665m/s^2

F = m*a = 0.15 * 2665 = 399.75 N.

To calculate the ball's acceleration and the force applied on it during contact with the bat, we can use Newton's second law of motion, which states that the force exerted on an object is equal to the product of its mass and acceleration:

Force = Mass * Acceleration

Step 1: Convert all the given velocities to SI units (m/s):

Initial velocity (v1) = 40.89 m/s
Final velocity (v2) = -46.22 m/s (opposite direction)

Step 2: Calculate the change in velocity:

Δv = v2 - v1 = -46.22 m/s - 40.89 m/s = -87.11 m/s

Step 3: Calculate the acceleration:

Acceleration = Δv / time

Given that the time (t) is 0.002 seconds:

Acceleration = -87.11 m/s / 0.002 s = -43,555 m/s^2

The negative sign indicates that the ball's velocity is decreasing, as it is being hit back towards the pitcher.

Step 4: Calculate the force applied on the ball during contact:

Force = Mass * Acceleration

Given that the mass of the ball (m) is 150 grams (0.15 kg):

Force = 0.15 kg * -43,555 m/s^2 = -6,532.25 N

The negative sign indicates that the force is acting in the opposite direction of the ball's initial motion.

So, during contact with the bat, the ball experiences an acceleration of approximately -43,555 m/s^2 and a force of approximately -6,532.25 N.