Typically, when you look at a real world examples for using a quadratic equation/formula, you get a ball being tossed or a missile being launched. I understand what each component of the equation stands for. However, my question is what angle is the object being thrown? I would think that the angle of trajectory would change the height and distance the object reaches. Nevertheless, all the real world examples never talk about this and use the same set up no matter if its a ball or missile being thrown. As such, is there a typical/average angle these real world examples are using?
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You can use whatever angle you want when you throw a ball. That's all I can say. – 3x89g2 Sep 13 '16 at 17:03
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For furthest distance, they choose 45 degrees. For highest altitude they just shoot the projectile straight up. Other angles come in when simplicity in the firing mechanism is desired. – Parcly Taxel Sep 13 '16 at 17:06
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Please edit the question to literally show one of the typical "real world examples" that you are talking about, including the equation that is used. There are plenty of examples of trajectory calculations in which an angle is explicitly stated, but you apparently are looking mostly at different examples. – David K Sep 13 '16 at 20:04
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Your equation only applies to the altitude (y-coordinate) of the object. You need a separate equation for the horizontal component (x-coordinate). With both equations, you can figure out the angle. If $x$ and $y$ are functions of time, $y=y(t)$ and $x = x(t)$, then the angle is the inverse tangent of $$ \dfrac {dy}{dx} = \dfrac{dy/dt}{dx/dt}$$
Steven Alexis Gregory
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Typically, such a problem will state a particular initial angle at which the projectile is thrown. Sometimes a problem will say "vertically", so the angle with the horizontal is 90 degrees, sometimes "horizontally" so the angle is 0 degrees.
user247327
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