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## Homework Statement

A batter hits two identical baseballs with the same initial speed and from the same initial height but at different initial angles. Prove that both balls have the same speed at any height h if air resistance can be neglected.

This was an example in the book and was "proven" completely conceptually. I thought it would be very interesting to tackle it mathematically. I probably have a mistake somewhere, and I'm not very familiar with proofs at all so bare with me. I'm 99% sure my attempt at the proof is non conclusive, help will be very appreciated!

## Homework Equations

[tex]w_{grav}=mgy_1 - mgy_2[/tex]

[tex]w_{grav}=U_{grav1}-U_{grav2}=-Δu[/tex]

[tex]u_{grav} = mgy[/tex]

[tex]\frac{1}{2}mv_{1}^2+mgy_1 = \frac{1}{2}mv_{2}^2+mgy_2[/tex]

## The Attempt at a Solution

[tex]v_{01}=v_{02}=v_1[/tex]

[tex]h_1=h_2=y[/tex]

[tex]m_1=m_2=m[/tex]

For baseball 1

[tex]\frac{1}{2}mv_{1}^2+mgy_1 = \frac{1}{2}mv_{2}^2+mgy_2[/tex]

[tex]\frac{\frac{1}{2}mv_{1}^2+mgy_1}{2} = \frac{\frac{1}{2}mv_{2}^2+mgy_2}{2}[/tex]

[tex]mv_{1}^2=mv_{2}^2 + \frac{Δu}{2}[/tex]

[tex]\frac{mv_{1}^2}{m} = \frac{mv_{2}^2 + \frac{Δu}{2}}{m}[/tex]

[tex]v_{1}^2=v_{2}^2+\frac{Δu}{2m}[/tex]

[tex]v = \sqrt{v_2^2+\frac{Δu}{2m}}[/tex]

Now let take into account baseball 2

[tex]\frac{1}{2}mv_{1}^2+mgy_1 = \frac{1}{2}mv_{2}^2+mgy_2[/tex]

[tex]\frac{1}{2}mv_{1}^2 - \frac{1}{2}mv_{2}= mgy_2-mgy_1[/tex]

where it implies

[tex]-Δk = Δu[/tex]

so from previous equation

[tex]v = \sqrt{v_2^2+\frac{Δk}{2m}}[/tex]

I can go on to simplify this more but it wouldn't do us any good. Anyhow the equation implies that final velocity is in a relationship with no variables holding change in y or angle. Therefore both baseballs will have the same speed at any height y at any angle.

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