What is the final velocity of the baseball after being hit by the bat?

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Homework Help Overview

The discussion revolves around a physics problem involving a baseball being hit by a bat. The original poster presents a scenario where a baseball is thrown at a certain speed and work is done on it by the bat. The goal is to determine the final velocity of the baseball after it has been hit and has risen to a specific height.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the application of the work-energy principle, questioning the initial setup and the use of gravitational acceleration in their calculations. There are attempts to clarify the relationship between initial kinetic energy and the work done by the bat.

Discussion Status

Some participants are exploring the conservation of energy approach, while others are questioning the assumptions made regarding the initial conditions and the equations used. There is a recognition of confusion regarding the application of projectile motion concepts in this context.

Contextual Notes

Participants note the lack of clarity on how to incorporate projectile motion without angles and express uncertainty about the initial kinetic energy calculations. There is also mention of the need for further demonstration of the concepts involved.

agadag
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Homework Statement


A pitcher throws a 0.124-kg baseball, and it approaches the bat at a speed of 54.2 m/s. The bat does Wnc = 76.5 J of work on the ball in hitting it. Ignoring the air resistance, determine the speed of the ball after the ball leaves the bat and is 21 m above the point of impact.


Homework Equations



W = E - E0

The Attempt at a Solution


W = E -Eo
76.5=.5(.124)v2-.124(9.8)21
102.0192=.062v2
v = wrong answer!
Please help. I don't have a basics in physics, so please explain from an beginners standpoint.
Thankyou!
 
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agadag said:

The Attempt at a Solution


W = E -Eo
76.5=.5(.124)v2-.124(9.8)21
102.0192=.062v2
v = wrong answer!

where did you get 9.8 from? It told you 54.2 m/s was the initial velocity. Redo this part. For the second part you need to use projectile motion equations.
 
I used the eqn W=E-E0, since they gave us a non conservative value.
this can b furthere broken down to... 1/2mv2 -mgh. Thats where the 9.8 comes from. As far as projectile motion..how would i use it with no angles?
 
agadag said:
I used the eqn W=E-E0, since they gave us a non conservative value.
this can b furthere broken down to... 1/2mv2 -mgh. Thats where the 9.8 comes from. As far as projectile motion..how would i use it with no angles?

actually as I think about it now, you don't need to use projectile motion.

Initially

all you have is that that change in kinetic energy = work done by that bat.

the initial ke = 1/2m(54)2.

For the second part you need to use the part in red
 
Im confused. I got KE = 180.792. however, I am trying to find final velocity of the ball. i initially tried the eqn in red and it did not work for me. Can you please demonstrate?
 
agadag said:
Im confused. I got KE = 180.792. however, I am trying to find final velocity of the ball. i initially tried the eqn in red and it did not work for me. Can you please demonstrate?

Conservation of energy right

Initially it has ke given by 1/2mu2. This energy is converted into kinetic energy (it is moving with a different velocity v) and work done by the bat.

so applying the law of conservation of energy we will get 1/2mu2=1/2mv2+Wbat.

So can you find 'v' given Wbat=76.5J and u=54.2m/s?
 

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