A baseball (m = 135 g) approaches a bat horizontally at a speed of 40.4 m/s (90 mi/h) and is hit straight back at a speed of 45.7 m/s (102 mi/h). If the ball is in contact with the bat for a time of 1.00 ms, what is the average force exerted on the ball by the bat? Neglect the weight of the ball, since it is so much less than the force of the bat. Choose the direction of the incoming ball as the positive direction.

a=(V-Vo)/t\=(45.7-40.4)/10^-3=5300 m/s^2

F = m*a = 0.135 * 5300 = 715.5 N.

To calculate the average force exerted on the ball by the bat, we can use Newton's second law of motion, which states that the force acting on an object is equal to the rate of change of its momentum.

Momentum (p) is defined as the product of an object's mass (m) and its velocity (v): p = m * v

In this case, the ball approaches the bat horizontally, so the initial momentum of the ball can be calculated as:

p_initial = m * v_initial

Given that m = 135 g = 0.135 kg and v_initial = 40.4 m/s, we have:

p_initial = 0.135 kg * 40.4 m/s

Now, let's calculate the final momentum of the ball. Since the ball is hit straight back, its final velocity (v_final) will be in the opposite direction of the initial velocity. So:

v_final = -45.7 m/s

The final momentum of the ball is:

p_final = m * v_final

p_final = 0.135 kg * (-45.7 m/s)

Next, we can calculate the change in momentum (Δp) by subtracting the initial momentum from the final momentum:

Δp = p_final - p_initial

Finally, using the fact that force (F) is the rate of change of momentum over time, we can calculate the average force exerted on the ball by the bat:

F = Δp / Δt

Before calculating the force, we need to convert the given time of 1.00 ms to seconds:

Δt = 1.00 ms = 0.001 s

Now, we can plug in the values and calculate the average force:

F = (Δp) / (Δt)

Note: To convert mi/h to m/s, multiply the value by 0.44704.

I will now calculate the average force using the given values.