Assume that a pitcher throws a baseball so that it travels in a straight line parallel to the ground. The batter then hits the ball so it goes directly back to the pitcher along the same straight line. Define the direction the pitcher originally throws the ball as the x direction.

v_f = -25.9 m/s

Explanation:

- The complete question is as follows:

" Assume that a pitcher throws a baseball so that it travels in a straight line parallel to the ground. The batter then hits the ball so it goes directly back to the pitcher along the same straight line. Define the direction the pitcher originally throws the ball as the +x direction.

Now assume that the pitcher in Part D throws a 0.145-kg baseball parallel to the ground with a speed of 32 m/s in the +x direction. The batter then hits the ball so it goes directly back to the pitcher along the same straight line. What is the ball's velocity just after leaving the bat if the bat applies an impulse of −8.4N⋅s to the baseball?"

Given:

- mass of baseball m = 0.145 kg

- Speed before impact v_i = 32 m/s

- Speed after impact v_f

- Impulse applied by the bat I = - 8.4Ns

Find:

What is the ball's velocity just after leaving the bat

Solution:

- Impulse is the change in linear momentum of the ball according to Newton's second law of motion:

                              I = m* ( v_f - v_i )

- Taking the + from pitcher to batsman and - from batsman to pitcher.

- Plug in the values:

                              -8.4 = 0.145* ( v_f - (32) )

                               v_f = -57.93103 + 32

                               v_f = -25.9 m/s

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