# python question.

*label*Science

*timer*Asked: Mar 7th, 2019

*account_balance_wallet*$25

### Question Description

This problem has two parts: the first part (the most important part) is solving the **final **nonlinear equation in the problem statement for ๐.

Once ๐ is known, you can this solve for the location of the football, i.e., solve for ๐ฅ and ๐ฆ using a series (or vector) of time values. I used numpy.linspace() function to create a time vector and then solved for ๐ฅ and ๐ฆ at each time point, i.e.,

time = numpy.linspace(0,4)

x = x0 + v0*math.cos(theta)*time

y = y0 + v0*math.sin(theta)*time - (g*time**2)/2.0

Then, you can generate an ๐ฅ vs. ๐ฆ plot showing the trajectory of the ball. If the ball doesnโt reach the target or if the ball goes through the ground (i.e., negative values for ๐ฅ) then you know that you choose a range of time values that is too short or too long. If youโve watched much (American) football, then you know that the ball is typically in the air for a few (more than 1 and less than 5 seconds) even for long throws.

Please make sure to include the input code and output as well as the memo. Please do it in your own work and words. Thanks

### Unformatted Attachment Preview

## Tutor Answer

Please take the code from this word file. PDF is making all the indentation disappear.

Football Throw Problem

First Part โ Determination of the throw angle

In the given final equation after equating the time taken to reach the final x and y

coordinates, the final equation becomes โ

๐. (๐ โ ๐๐ )๐

๐๐ + ๐ญ๐๐ง ๐ฝ . (๐ โ ๐๐ ) โ

โ๐=๐

๐. ๐๐๐ . ๐๐๐๐ ๐ฝ

Now, using the following relationships and changing the terms in the above equation will

give us โ

Relationships and terms used

1. The change in the x-coordinates are denoted as delta-x, denoted by ฮx.

โ๐ฅ = ๐ฅ โ ๐ฅ0

2. The change in the y-coordinates are denoted as delta-x, denoted by ฮy.

โ๐ฆ = ๐ฆ โ ๐ฆ0

3. The value of cos ฮธ can be re written as (1 / sec ฮธ) and using the relationship between

tan ฮธ and sec ฮธ, we get โ

1

= sec 2 ๐ = tan2 ๐ + 1

2

cos ๐

4. And considering the initial velocity and acceleration due to gravity to be constants,

we write a new constant K

๐พ=

๐

2. ๐ฃ20

The Final equation becomes โ

K. (โx)2 (tan2 ๐ + 1) โ โx. tan ๐ + โy = 0

This is clearly a quadratic equation of the (tan ฮธ), which gives two values of ฮธ.

Conclusion: These two values point to two possible angles, and a flat throw and a

lobbed-up throw, with same initial velocity and same final distance.

Using the Python code (Appendix - A) using Broydenโs (Broyden1) method and taking

two different initial guesses, we have found those two values of the ฮธ to be โ

Measurement

ฮธ1 (Flat Throw, ...

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