A baseball pitcher can throw a baseball at a velocity of 44.5 m/s. If the mass of the ball is 0.12 kg and the pitcher has it in his hand over a distance of 1.79 m, what is the average force exerted by the pitcher on the ball?

a = (Vf^2 - (Vo)^2) / 2d,

a = ((45)^2 - 0) / 3.58 = 565.6m/s^2.

F = ma = 0.12 * 565.6 = 67.9N.

To find the average force exerted by the pitcher on the ball, we can use Newton's second law of motion, which states that force (F) is equal to the mass (m) of an object multiplied by its acceleration (a), or F = m * a.

In this case, we know the mass of the ball (m = 0.12 kg) and we can find the acceleration by dividing the change in velocity (Δv) by the time taken (Δt). However, since we are given the distance covered by the ball (d), we need to find the time taken (Δt) using the formula for average velocity, which is v = d / t.

Given:
- Velocity (v) = 44.5 m/s
- Mass (m) = 0.12 kg
- Distance (d) = 1.79 m

First, let's find the time taken (Δt):
v = d / t
44.5 m/s = 1.79 m / t

Next, solve for t by rearranging the equation:
t = 1.79 m / 44.5 m/s
t ≈ 0.0402 s

Now we have the time taken (Δt), we can find the acceleration (a):
a = Δv / Δt
Since the initial velocity is 0 m/s (the ball is at rest in the pitcher's hand), the change in velocity (Δv) is equal to the final velocity (v).
a = v / t
a = 44.5 m/s / 0.0402 s
a ≈ 1106.97 m/s²

Finally, calculate the average force exerted by the pitcher on the ball using Newton's second law of motion:
F = m * a
F = 0.12 kg * 1106.97 m/s²
F ≈ 132.83 N

Therefore, the average force exerted by the pitcher on the ball is approximately 132.83 Newtons.