Question

13. A baseball pitcher throws a fastball at a speed of 46m/s. The
acceleration occurs as the pitcher holds the ball in his hand and moves it
through an almost straight line distance of 3.5 m. Calculate the
acceleration, assume it is constant and uniform. (Answer in 3 sig figs and
do not include units) *

Answer 1

Answer: Approximately 302 m/s^2

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Work Shown:

s = starting velocity = 0

f = final velocity  = 46

d = distance = 3.5

a = acceleration = unknown (we're solving for this)

$f^2 = s^2 + 2a*d \ \ \text{ ..... one of the kinematics equations}\\\\46^2 = 0^2 + 2a*3.5\\\\2116 = 7a\\\\7a = 2116\\\\a = \frac{2116}{7}\\\\a \approx 302.28571\\\\a \approx 302$

The acceleration to three sig figs is roughly 302 m/s^2

The acceleration is so large because the ball's final velocity is incredibly fast in such a short amount of time.