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I have two equations on my equation sheet:

$$y=v_o t \sin \theta - \frac{1}{2} g t^2.$$

$$y_{\max} = \frac{v_o^2 \sin^2\theta }{2g}.$$

I understand that the first one is an equation for the height of the projectile (as a function of time $t$), but what is the second equation for?

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  • $\begingroup$ How do you find the maximum value of a function? $\endgroup$ – tpg2114 Apr 22 '14 at 23:58
  • $\begingroup$ But wouldn't that be the exact same thing as the first equation? It finds the height of the throw (maxima) $\endgroup$ – user3200098 Apr 23 '14 at 0:00
  • $\begingroup$ The first equation has time in it -- so it doesn't find the maximum, it finds the height as a function of time. $\endgroup$ – tpg2114 Apr 23 '14 at 0:00
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$$\Delta y = v_0 \sin(\theta) t - \frac12 gt^2$$

This is a projectile, so it will hit its max at $t_{max}=\frac12t$, where $t$ is the total time in which the projectile flies.

The total time is when, as you know, $\Delta y=0$. Hence we've got:

$$0=v_0 \sin(\theta) t - \frac12 gt^2$$ $$\require{cancel}\frac12 gt^\cancel{2}=v_0 \sin(\theta) \cancel{t}$$ $$t=\frac{v_0 \sin(\theta)}{\frac12 g}=\frac{2v_0 \sin(\theta)}{g}$$

That's the total time of the projectile motion. Take half of that to get the time of the maximum height:

$$\frac12 t = t_{max} = \frac{v_0 \sin(\theta)}{g}$$

Now plug it in back to $\Delta y$ and you'll get the maximum $y$ displacement:

$$\Delta y_{max} = v_0 \sin(\theta)t_{max} - \frac12 gt_{max}^2$$

$$\require{cancel}\Delta y_{max}= v_0 \sin(\theta)\left(\frac{v_0 \sin(\theta)}{g}\right) - \frac12 g\left(\frac{v_0 \sin(\theta)}{g}\right)^2=\frac{v_0^2 \sin^2(\theta)}{g} - \frac12 \cancel{g}\left(\frac{v_0^2 \sin^2(\theta)}{g^\cancel{2}}\right)=\frac{v_0^2 \sin^2(\theta)}{2g}$$

Just for bonus, here's a mathematical proof as to why it's at maximum height halfway through:

$$y(t)=v_0 \sin(\theta) t - \frac12 gt^2$$ $$\frac{dy}{dt}=y'(t)=v_0 \sin(\theta) - gt$$ $$0=v_0 \sin(\theta) - gt\Rightarrow gt=v_0 \sin(\theta) \Rightarrow t=\frac{v_0 \sin(\theta)}{g}$$

Which is, again, halfway. Of course it's a maximum because:

$$\frac{d^2y}{dt^2}=y''(t)=-g$$

Cheers! -Shahar

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