Question

a baseball player pitches a ball at 40.9 m/s. if a standard baseball has a mass of 0.146 kg and it is in contact with the hand for 0.128 s, how much power does he exert on the ball?

320 w
18.9 w
954 w
481 w

Explanation:

Identify given values and target

We identify the given physical parameters:

• Initial velocity \(v_i = 0\text{ m/s}\)
• Final velocity \(v_f = 40.9\text{ m/s}\)
• Mass of the baseball \(m = 0.146\text{ kg}\)
• Time interval \(\Delta t = 0.128\text{ s}\)

Our target is to find the average power \(P\) exerted on the ball.

Calculate kinetic energy change

The work done on the ball equals its change in kinetic energy:

$$ W = \Delta KE = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 $$

$$ W = \frac{1}{2} (0.146\text{ kg}) (40.9\text{ m/s})^2 $$

$$ W = 0.073 \times 1672.81 \approx 122.115\text{ J} $$

Calculate average power

Power is the rate at which work is done over time:

$$ P = \frac{W}{\Delta t} $$

$$ P = \frac{122.115\text{ J}}{0.128\text{ s}} \approx 954.02\text{ W} $$

Match with options

The calculated power is approximately \(954\text{ W}\), which matches the third option.

Answer:

• 320 W
• 18.9 W
• 954 W (Correct answer)
• 481 W