In 1947 Bob Feller, a former Cleveland pitcher, threw a baseball at 98.6 mph or 44.1 m/s.
For maney years this was the fastest pitch ever measured. If Bob had thrown the ball straight up,
how high would it have gone?

Other information that is not explicitly stated in the problem, but that we know

Check the given information (If Possible)

We need to Find the Height

The intial velocity of the ball is 44.1 m/s, Straight up

We can assume the following:

The acceleration due to gravity on earth
is constant, acts downward, will slow the ball to 0 m/s
at the top of its arc, and has a value of a_{g} = 9.81 m/s^{2} Down.

At maximum height the ball momentarily stops,
before falling back to earth.

We will ignore Air Resistance.

We Should check to see that the 98.6 mph has been correctly converted

Step 2) Describe the problem in terms of the field

What does this have to do with ...?

Here we are seeking to connect all the pieces to see the relationships

We will want to include the following:

A well labeled Diagram

A verbal description of what is happening in the problem.

The above can often be complemented with plots that help to describe the underlying physics.

Drawing: Vertically Thrown Ball

Verbal Decsription

In this case the intial velocity V_{i} decreases linearly with
respect to time due to the graviational acceleration. The maximum height H occurs
at the top, when the velocity is zero and the time is given by t_{m}

Descriptive Plot

In this case a plot of what happens to the velocity of
the raising ball can be very helpfull.

As we can see the graph starts at 44.1 m/s (y - intercept) and has a downward slope
that matches the acceleration due to gravity. From the graph we can see that the time
to the top, which occurs when the velocity reaches 0 m/s, is roughly 4.5 seconds
This will be helpful later on in checking our work.

Step 3) Plan a Solution

How do I get out of this?

Here we are seeking to connect all the pieces to see the relationships

We will want to include the following:

Using the following:
V_{i} = 44.1 m/s
V_{m} = 0 m/s - Velocity at the top of the path
a_{g} = -9.81 m/s^{2}

Using two of our motion equations

V_{f} = V_{i} + at

D = V_{i}t + (1/2)at^{2}

We will need to find the time to the top t_{m} and then use that
result to find H, the maximum height.
Step 4) Execute the Plan

Lets Get an Answer

First we will solve for the time and then use that result
to calculate the height
V_{i} = 44.1 m/s
V_{m} = V_{f} = 0 m/s - Velocity at the top of the path
a_{g} = a = -9.81 m/s^{2}

V_{f} = V_{i} + at
V_{m} = V_{i} + a_{g}t_{m}
0 = 44.1 -9.81t_{m}
t_{m} = 4.5 seconds
Now we can find the height using:
D = V_{i}t + (1/2)at^{2}
H = V_{i}t_{m} + (1/2)a_{g}(t_{m})^{2}
H = 44.1(4.5) + (1/2)(-9.81)(4.5)^{2}
H = 99.1 m

Step 5) Evalute the Solution

Can This be True

Our calculated t_{m} and the value obtained from the plot are identical

The ball was thrown very fast so we would expect it to rise quite high

We discounted air resistence

Therfore since we did not make any mathmatical miscalculations in the last part of
our solution, we can conclude that our solution is valid