# Kinetic energy of a Baseball

1. Sep 26, 2009

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

W = E - E0

3. The attempt at a solution
W = E -Eo
76.5=.5(.124)v2-.124(9.8)21
102.0192=.062v2
Thankyou!!!

2. Sep 26, 2009

### rock.freak667

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.

3. Sep 26, 2009

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?

4. Sep 26, 2009

### rock.freak667

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

5. Sep 26, 2009

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?

6. Sep 26, 2009

### rock.freak667

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?