# Kinetic Energy of a Baseball

1. Feb 21, 2008

### cstout

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

A pitcher throws a 0.128-kg baseball, and it approaches the bat at a speed of 56.1 m/s. The bat does Wnc = 81.3 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 24.6 m above the point of impact.

2. Relevant equations

Change in Linear Momentum, Linear Velocity, Work

3. The attempt at a solution

2. Feb 21, 2008

### ||spoon||

well what have you got so far?

3. Feb 21, 2008

### cstout

well i know that you have to find the final velocity at d=24.6m but I'm not sure how to fine the other variables that go into that equation.

4. Feb 21, 2008

### ||spoon||

it may be a good to find the energy of the ball the moment before it is hit by thebat for a starting point ;)

5. Feb 21, 2008

### cstout

ok i solved for that and got 1573.669, what would the next step be

6. Feb 21, 2008

### rohanprabhu

is this the energy? Please use proper units when conveying the answer.

Now, that u have the energy, and u know the work the bat does. By law of conservation of energy, the total energy i.e. the energy u found out + the work done by the bat must be converted to the kinetic energy of the ball. Now, can u find the velocity with this?

7. Feb 21, 2008

### cstout

yea the total energy plus the work done by that bat is 1654.969J and after putting that into the kinetic energy equation of 1654.969=.5(.128)v^2, I got v=160.81, but the question wants velocity at 24.6m from the point of impact

8. Feb 21, 2008

### ||spoon||

forget getting the velocity just yet. If you use the energy plus the work of the ball which you found, subtracting from this the gravitational potential energy the ball gains by rising 24.6 m (using mgh). What energy is left is kinetic energy. Now you can find the velocity.

9. Feb 21, 2008

### rohanprabhu

I am getting the total work as 282.72 J. And v = 66.46..

Now, since you want the velocity of the ball 24.6m *above* the point of impact.. i'm gonna assume that the hitter hit the ball straightway upwards, perpendicular to the ground. In this case an acceleration of $g = 9.8 ms^{-2}$ is acting on it. Use the formula:

$$v^2 - u^2 = 2as$$

where, u = initial velocity [which we calculated], v = final velocity [which is required], a = acceleration [in our case, the acceleration due to gravity] and s = distance travelled.

Also, pay attention to the signs you assign to these variables.

Last edited: Feb 21, 2008