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On this site I've found a formula for calculating the $x, y$ coordinates for a body throwed by an angle to a horizon.

It looks like this:
$$x(t) = V_0 t \cos(\alpha); $$ $$y(t) = V_0 t \sin(\alpha) - gt^2/2; $$

It's clear to me that term $(g t^2)/2$ was received by the integration of $gt \, dt$.
But, I don't understand, why didn't we integrate $V_0 t \sin(\alpha)$ part of the equation?
Why do we integrate only $gt$ term?

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  • $\begingroup$ Everything were integrated. That's why there is a $t$ in $v_0 t\sin{a}$. $\endgroup$ – Physicist137 Nov 19 '14 at 23:51
  • $\begingroup$ @Physicist137, so, the original equation was then x(t)=V0*sin(a), is that's right? $\endgroup$ – PaulD Nov 20 '14 at 0:06
  • $\begingroup$ Well, no. You must integrate velocity to get position. Hence, original equation was: $v_y(t) = v_0\sin a - gt$ and $v_x = v_0\cos a$. $\endgroup$ – Physicist137 Nov 20 '14 at 0:14
  • $\begingroup$ @Physicist137, now I understand, thanks. $\endgroup$ – PaulD Nov 20 '14 at 0:22
  • $\begingroup$ $\int V_0 sin(\alpha) dt = V_0 sin(\alpha) t$ $\endgroup$ – Jold Nov 20 '14 at 0:36

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