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 x-component of velocity just after leaving the bat if the bat applies an impulse of −8.4n⋅s to the baseball

Explanation:

Given that,

Mass of the baseball, m = 0.145 kg

Initial speed of the baseball, u = 32 m/s

Impulse applied by the bat to the baseball, J = -8.4 N-s

To find,

The ball's x-component of velocity just after leaving the bat.

Solution,

The impulse imparted from one object to other is equal to the change in momentum. Mathematically, it is given by :

On solving the above equation, we can find the value of v as :

Therefore, the velocity of the ball just after leaving the bat is -25.931 m/s. Hence, this is the required solution.

The ball's x-component of velocity just after leaving the bat is -25.93 m/s.

Given data:

The mass of baseball is, .

The speed of throwing baseball in x-direction is, .

The impulse produced by the batter is,

Apply the impulse-momentum equation as,

Here, v is the final speed of baseball after hitting.

Solving as,

Negative sign shows that its direction is opposite to  speed of throwing.

Thus,  the ball's x-component of velocity just after leaving the bat is -25.93 m/s.

Learn more about concept of impulse-momentum theorem here:

link

im not to sure what it is

no. dot product   =0 when vecs at right angles