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So let's say I have a parabolic function that describes the displacement of some projectile without air resistance. Let's say it's

$$y=-4.9x^2+V_0x.$$

I want to know at what angle the projectile was fired. I notice that

$$\tan \theta_0=f'(x_0)$$ so the angle should be

$$\theta_0 = \arctan(f'(0)).$$ or

$$\theta_0 = \arctan(V_0).$$

Is this correct? I can't work out why it wouldn't be, but it doesn't look right when I plot curves.

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    If you intend -9.8 to be the acceleration of gravity, the altitude it contributes should be $\frac {-9.8}2 x^2$ (where you are using $x$ for time, not horizontal distance)2012-08-28
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    oops! Yep. I forgot. :)2012-08-28
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    As the others say, if $x$ is time you are working in one dimension and there is no $\theta_0$ If you are in two dimensions, you need $x$ and $t$.2012-08-28
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    @Ross: If the OP meant $x$ to be time, you're correct. However, it seems more likely to me that they really did mean $x$ to be horizontal position, in which case $\theta=\arctan f'(x_0)$ does, in fact, give the angle of the trajectory (up to sign, anyway) at position $(x_0,f(x_0))$.2012-08-29
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    @IlmariKaronen: but then the first equation doesn't work in units at all. The acceleration of gravity shouldn't multiply distance^2 to get distance. OP edited in response to my first comment, which seems to confirm that $x$ is time.2012-08-29

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