Projectile Motion of Soccer Ball Problem

[Precursor]

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

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

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.

Ax and Ay are the horizontal and vertical components of the initial velocity respectively.

14 = \sqrt{Ax^{2} + Ay^{2}}
14^{2} = Ax^{2} + Ay^{2}
Ax^{2} = 14^{2} - Ay^{2}

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

12.9 = \sqrt{Ax^{2} + By^{2}}
12.9^{2} = Ax^{2} + By^{2}
Ax^{2} = 12.9^{2} - By^{2}

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

14^{2} - Ay^{2} = 12.9^{2} - By^{2}

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

Ay = By + (9.81)(0.275)
By = Ay - 2.69775

Now, we can proceed:

196 - Ay^{2} = 166.41 - (Ay - 2.69775)^{2}
196 - Ay^{2} = 166.41 - Ay^{2} + 5.3955Ay - 7.2779
Ay = 6.833 m/s

Finally, we can solve for the angle:

\theta = sin^{-1} (Ay/14)
\theta = sin^{-1} (6.833/14)
\theta = 29.2^{o}

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.

 

[ehild]

Ay = 6.83 m/s, that is right. But Ay/14 is the sine of the angle, the tangent is Ay/Ax.

ehild

 

[Precursor]

Ay = 6.83 m/s, that is right. But Ay/14 is the sine of the angle, the tangent is Ay/Ax.

ehild

Yeah that's what I actually had written in my solution. I just transferred it wrong. I'll fix it.