Answer:
[tex]v=-25.931\ m/s[/tex]
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 :
[tex]J=m(v-u)[/tex]
[tex]-8.4=0.145(v-32)[/tex]
On solving the above equation, we can find the value of v as :
[tex]v=-25.931\ m/s[/tex]
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, [tex]m=0.145 \;\rm kg[/tex].
The speed of throwing baseball in x-direction is, [tex]u=32 \;\rm m/s[/tex].
The impulse produced by the batter is, [tex]I=-8.4 \;\rm N-s[/tex]
Apply the impulse-momentum equation as,
[tex]I=m(v-u)[/tex]
Here, v is the final speed of baseball after hitting.
Solving as,
[tex]-8.4=0.145 \times(v-32)\\-57.93=v-32\\v=-25.93 \;\rm m/s[/tex]
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:
https://brainly.com/question/14121529?referrer=searchResults
