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## Homework Statement

A soccer ball is kicked from the ground with an initial speed of 14.0 m/s. After 0.275 s its speed is 12.9 m/s. Find the ball's initial direction of motion.

## Homework Equations

[tex]\theta = sin^{-1} (y/h)[/tex]

## The Attempt at a Solution

The work that I show below may not be very clear because I did not really use any physics equation. It was more math than anything, but I will do my best.

[tex]Ax[/tex] and [tex]Ay[/tex] are the horizontal and vertical components of the initial velocity respectively.

[tex]14 = \sqrt{Ax^{2} + Ay^{2}}[/tex]

[tex]14^{2} = Ax^{2} + Ay^{2}[/tex]

[tex]Ax^{2} = 14^{2} - Ay^{2}[/tex]

Below, I am using [tex]Ax[/tex] again, because the horizontal component is not impeded by anything as the projectile proceeds(assuming no resistance).

[tex]12.9 = \sqrt{Ax^{2} + By^{2}}[/tex]

[tex]12.9^{2} = Ax^{2} + By^{2}[/tex]

[tex]Ax^{2} = 12.9^{2} - By^{2}[/tex]

As you can see above, I have isolated [tex]Ax[/tex] so that I can equate the two equations, as shown below:

[tex]14^{2} - Ay^{2} = 12.9^{2} - By^{2}[/tex]

Above we have two unknown variables. Below I will relate them so that I can proceed with the above equation.

[tex]Ay = By + (9.81)(0.275)[/tex]

[tex]By = Ay - 2.69775[/tex]

Now, we can proceed:

[tex]196 - Ay^{2} = 166.41 - (Ay - 2.69775)^{2}[/tex]

[tex]196 - Ay^{2} = 166.41 - Ay^{2} + 5.3955Ay - 7.2779[/tex]

[tex]Ay = 6.833 m/s[/tex]

Finally, we can solve for the angle:

[tex]\theta = sin^{-1} (Ay/14)[/tex]

[tex]\theta = sin^{-1} (6.833/14)[/tex]

[tex]\theta = 29.2^{o}[/tex]

So my final answer is 29.2 degrees. Can someone please verify this answer? I feel somewhat skeptical about my answer because my method was quite unconventional.

Thank you.

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