# AP Physics 1

## Problem Solving

### Example:

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?

Step 1) Recognize the Problem

#### What is going On?

Here we need to consider the following:

1. What does the problem ask for?
2. What information is Given?
3. Other information that is not explicitly stated in the problem, but that we know
4. 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:
1. 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 ag = 9.81 m/s2 Down.
2. At maximum height the ball momentarily stops,
before falling back to earth.
3. 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:

1. A well labeled Diagram
2. 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 Diagram of Ball being thrown straight up

### Verbal Decsription

In this case the intial velocity Vi 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 tm

### 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. Velocity of a Ball Affected by Gravity
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:
Vi = 44.1 m/s
Vm = 0 m/s - Velocity at the top of the path
ag = -9.81 m/s2

Using two of our motion equations

• Vf = Vi + at
• D = Vit + (1/2)at2

We will need to find the time to the top tm 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
Vi = 44.1 m/s
Vm = Vf = 0 m/s - Velocity at the top of the path
ag = a = -9.81 m/s2

Vf = Vi + at
Vm = Vi + agtm
0 = 44.1 -9.81tm
tm = 4.5 seconds
Now we can find the height using:
D = Vit + (1/2)at2
H = Vitm + (1/2)ag(tm)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 tm 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