Projectile Range Symmetry Proof

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A projectile launched at angle @ has the same horizontal range as one launched at angle (90 - @) when both are fired with the same initial speed. The discussion revolves around using kinematic equations to prove this symmetry. The initial velocity is decomposed into vertical (V_y) and horizontal (V_x) components, where V_y = V sin(@) and V_x = V cos(@). The relationship sin(90 - @) = cos(@) is noted, leading to the conclusion that both angles yield the same range due to their complementary nature. Understanding this symmetry is crucial for solving projectile motion problems effectively.
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Homework Statement



Prove that a projectile launched at angle @ has the same horizontal range as one launched with the same speed at angle (90 - @).

Homework Equations



Obviously kinematics will be used =/

The Attempt at a Solution



I've been messing around with kinematics but haven't gotten anywhere. I also know that sin(90-@) = cos(@) . Help appreciated :]
 
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let the initial speed be V, and decomposite into V_y=Vsin@ and V_x=Vcos@. Now use this to first calculate the time the projectile is in the air and then the range of the projectile.
 
I appreciate the help but when you decomposite the V wouldn't it be:

V_y = sin(90-@)
V_x = cos(@)
 
uhm no, you said yourself that sin(90-@)=cos@, so that would give V_y=V_x :)
 
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