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Football Throw Problem
First Part – 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 –
𝒈. (𝒙 − 𝒙𝟎 )𝟐
𝒚𝟎 + 𝐭𝐚𝐧 𝜽 . (𝒙 − 𝒙𝟎 ) −
𝟐. 𝒗𝟐𝟎 . 𝒄𝒐𝒔𝟐 𝜽
Now, using the following relationships and changing the terms in the above equation will
give us –
Relationships and terms used
1. The change in the x-coordinates are denoted as delta-x, denoted by Δx.
∆𝑥 = 𝑥 − 𝑥0
2. The change in the y-coordinates are denoted as delta-x, denoted by Δy.
∆𝑦 = 𝑦 − 𝑦0
3. The value of cos θ can be re written as (1 / sec θ) and using the relationship between
tan θ and sec θ, we get –
= sec 2 𝜃 = tan2 𝜃 + 1
4. And considering the initial velocity and acceleration due to gravity to be constants,
we write a new constant K
The Final equation becomes –
K. (∆x)2 (tan2 𝜃 + 1) − ∆x. tan 𝜃 + ∆y = 0
This is clearly a quadratic equation of the (tan θ), which gives two values of θ.
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 Broyden’s (Broyden1) method and taking
two different initial guesses, we have found those two values of the θ to be –
θ1 (Flat Throw, ...