# Trajectory Problem with Parabola

The height of an object is given by $$h(x) = -0.005x^2 + x.$$ When does the object hit the ground? When does it attain its maximum height? What is its maximum height?

I divided -.005/-1 to get 100, replaced x with 100, and ended up getting y = 50 when it hits the ground. That is what I got as the Range, when x = 0 (stop me if I'm wrong), or does -b/2a get me the maximum of the vertex? I'm, confusing myself now. If someone could give me a little explanation I would greatly appreciate it.

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How do you get 100 when you divide -.005 by -1? – Gerry Myerson Mar 14 '12 at 6:12
@AustinMohr Yes to your question, sorry I was not clear. – stytown Mar 14 '12 at 6:20

From your equation $h(x)=-0.005x^2+x$, we get that $h(x)=0$ when $x(-0.005x+1)=0$. This happens when $x=0$, and also when $-0.005x+1=0$. Rewrite this as $0.005x=1$. We get $x=\frac{1}{0.005}=200$.

The maximum occurs halfway between $x=0$ and $x=200$, that is, at $x=100$. Substitution shows that $h(100)=50$.

Another way of doing it is by using a remembered formula. And indeed if $a=-0.005$ and $b=1$, the maximum height happens when $x=\frac{-b}{2a}$, which in this case gives $x=100$.

Then you substitute in the height formula to get the maximum height. The $50$ that you get is the maximum height, and has nothing to do with hitting the ground. For that, we got $x=200$.

Remark: I will add some theory that may help in the long run. We have $h(x)=-0.005x^2+x$. Note that $0.005=\frac{1}{200}$. So we can rewrite the formula for $h(x)$ as $$h(x)=-\frac{1}{200}(x^2-200x).$$ Complete the square. We get $$h(x)=-\frac{1}{200}((x-100)^2 -10000)$$ or more simply $$h(x)=50-\frac{1}{200}(x-100)^2.$$ The term $\frac{1}{200}(x-100)^2$ is always $\ge 0$, and is $0$ when $x=100$. At that value of $x$, $h(x)$ is as big as it will ever get, namely $50$.

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Thanks, very clear explanation, exactly what I was looking for. – stytown Mar 14 '12 at 6:21
Double thanks for the even broader explanation! – stytown Mar 14 '12 at 6:29