(II) A baseball pitcher throws a baseball with a speed of 43 ms. Estimate the average acceleration of the ball during the throwing motion. In throwing the baseball, the pitcher accelerates it through a displacement of about 3.5 m, from behind the body to the point where it is released

The average acceleration of the ball during the throwing motion is 265.14m/s².

In order to get the acceleration, the Newton's third law of motion will be used. This will be:

v² = u² + 2as

We'll make a to be the subject of the formula and this will be:

a = (v² - u²) / 2s

We'll plug in the value into the equation and this will be:

a = (43² - 0) / (2 × 3.5)

a = 1849 / 7

= 264.14 m/s²

Therefore, the acceleration is 265.14m/s.

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a = 264.14 m/s²

Explanation:

From the question;

Initial velocity; u will be 0 m/s since the ball will start from rest.

Final velocity; v = 43 m/s

distance covered by the motion; s = 3.5m

To get the acceleration, we will make use of Newton's third equation of motion which is;

v² = u² + 2as

Making a the subject, we have;

a = (v² - u²)/2s

Plugging in the relevant values to give;

a = (43² - 0)/(2 × 3.5)

a = 264.14 m/s²

he kicked it 10 feet straight up into the air? i think this works.