How Fast is the Baseball After Being Hit 24.6m Up?

[cstout]

Homework Statement

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.

Homework Equations

Change in Linear Momentum, Linear Velocity, Work

The Attempt at a Solution

 

[||spoon||]

well what have you got so far?

 

[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.

 

[||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 ;)

 

[cstout]

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

 

[rohanprabhu]

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

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?

 

[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

 

[||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.

 

[rohanprabhu]

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

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 going to 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.