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I'm having problems to proof the equation for maximum height which is given as follows:

$$H_{\max}=\frac{v_o\sin^2\omega}{2\times g}$$

starting from here (which is the equation for $y$):

$$y=v_{oy}t-\frac{1}{2}gt^2$$

I am confused whether if the speed on $y$-axis becomes $0$ in the maximum height but would not this cancel the first term in the above equation?.

What would be the way to go?

Edit:

Although there can be different ways to assess this situation. What would be the one that explains what is happening to the vectors while the projectile is in the top?

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  • $\begingroup$ I gave the easiest. It is basic algebra. $\endgroup$ – David Peterson Dec 14 '17 at 14:24
  • $\begingroup$ @DavidPeterson I removed the "easiest" analytical geometry is fine but I'm looking for something which can explain my earlier question as the vertical vector becomes zero when the projectile is in the toppest. Wouldn't this make the $v_oy$ be zero in the stated function? $\endgroup$ – Chris Steinbeck Bell Dec 14 '17 at 14:31
  • $\begingroup$ You need calculus to "really" understand. $v(t)$ is zero when $y'(t)=0$. The rate of change of $f(t)=at^2+bt+c$ is given by $v(t)=2at+b$. If you set this equal to zero and solve, you'll find $t=-b/(2a)$ is when the velocity is zero. You're not plugging in $0$ for $v_{0y}$ that is the initial velocity only. $\endgroup$ – David Peterson Dec 14 '17 at 14:36
  • $\begingroup$ @DavidPeterson Thanks for clearing up my ideas. I can see that you are equating the derivate to zero as means to obtain the root. I'm just wondering if the observation that the vertical vector of the velocity has any use in the derivation of the answer other than the method you explained?. $\endgroup$ – Chris Steinbeck Bell Dec 14 '17 at 14:54
  • $\begingroup$ The velocity vector is given by $v=2at+b$, which in your case is $v=-gt+v_{0y}$ $\endgroup$ – David Peterson Dec 14 '17 at 15:24
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The vertex of the parabola occurs at $t=-\dfrac{b}{2a} =\dfrac{v_{0y}}{g}$. Substituting $v_{0y}=v_0\sin\omega$ and evaluating the function there should give you what you're after.

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  • $\begingroup$ As you pointed out. I reached to the "answer" (the maximum height), but I'm confused on what's the necessity of $t=-\frac{b}{2a}$? As I ommited this in the function. Or could it be that I'm doing it wrong? $\endgroup$ – Chris Steinbeck Bell Dec 14 '17 at 14:25
  • $\begingroup$ $f(x)=ax^2 + bx + c$ has vertex $(-b/(2a), f(-b/(2a)))$. The $y-$ coordinate is what you are after. $\endgroup$ – David Peterson Dec 14 '17 at 14:28
  • $\begingroup$ Yes I obtained the $y-coordinate$ by just following from what you solved but I got confused at why you mentioned the vertex, it turns out that the formula for the vertex was the way how to obtain the $x$, once this is computed the rest is evaluating the function as you already written. $\endgroup$ – Chris Steinbeck Bell Dec 14 '17 at 14:46
  • $\begingroup$ Are you saying you don't understand why finding the vertex gives you the maximum height of the projectile? Just graph the function and note where the maximum point and vertex are. $\endgroup$ – scott Dec 14 '17 at 20:50
  • $\begingroup$ @scott I understood the vertex is the maximum height and of course I have seen the graph. My question was where did the $\frac{-b}{2a}$ came from. After doing research on my own and from the answer that was given I figured out that it was related to the equation of the parabola which can be derived from its general form. Anyway thanks for showing interest in assisting. $\endgroup$ – Chris Steinbeck Bell Dec 18 '17 at 15:33

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