Solving for the interval of time during which the height of a thrown ball is at least “h” feet

In dealing with inequalities I've run into a certain peculiarity which I am currently unable to explain.

The example: Find the interval of time during which the ball is at least 32 feet above ground.

h = -16t^2 + 16t + 128 // Height of the ball in feet.

-16t^2 + 16t + 128 >= 32
-16t^2 + 16t + 96 >= 0
-16(t^2 - t - 6) >= 0
-16(t+2)(t-3) >= 0 // At this point everything is going as planned.


// Now I have a choice to make
(-16t - 32)(t - 3) >= 0 // This does not work  OR
(t + 2)(-16t + 48) >= 0 // This does work


// If I choose option 1, the relational operators are incorrect.
-16t - 32 >= 0      and      t - 3 >= 0
-16t >= 32                   t >= 3
t <= -2


// If I choose option 2, the relational operators are correct.
t + 2 >= 0          and      -16t + 48 >= 0
t >= -2                      -16t >= -48
t <= 3


Now, when looking at a graph it becomes obvious that the ball is at least 32 feet above ground during the interval [0, 3] (assuming time is not negative). Therefore, option 2 provides the correct relations for t, while option 1 inverts the relations.

What I don't understand is why this is happening, since multiplication is an associative operator. Seems to me that it shouldn't matter whether the -16 is multiplied into the first factor or the second, and yet it does. I would love to know why, so that I might circumvent this issue the next time around.

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I appreciate how you fully wrote out your work and showed your progress. Thank you. –  mixedmath Jul 19 '11 at 4:04

It comes from your line where you require both

\begin{align} -16t - 32 &> 0 \\ t - 3 &> 0 \end{align}

As you saw, this is not sensible. But you might instead require both

\begin{align} -16t - 32 &< 0 \\ t - 3 &< 0 \end{align}

And this does lead to the correct answer. Recall that the final answer is when either of these two cases are true.

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In other words, Bradford, if a product is positive, it might be that both factors are positive, but it might also be that both are negative. –  Gerry Myerson Jul 19 '11 at 4:43
@mixedmath, thank you for getting me back on track. I had forgotten that this approach was not useful in solving inequalities (but rather was useful only as a check). –  Bradford Fisher Jul 19 '11 at 5:04
@Gerry Myerson, thank you, I do see this now. I was so focused on the little details that I forgot to consider the equation as a whole. –  Bradford Fisher Jul 19 '11 at 5:15

Absorbing the $-16$ into one of the terms was unnecessary, as we will see later. But let's start from your actual calculation.

Look at the version that "doesn't work", namely $$(-16t-32)(t-3) \ge 0.$$ I have rewritten $\lt$ as $\le$, and $\gt$ as $\ge$, because "at least $32$ feet" means $32$ or more.

The displayed inequality is true if (i) $-16t-32$ and $t-3$ are both $\ge 0$ OR (ii) $-16t-32$ and $t-3$ are both $\le 0$.

Note that $-16t-32 \ge 0$ iff $-16t \ge 32$ iff $t\le -2$.

The condition $t-3\ge 0$ can be rewritten as $t\ge 3$. So our analysis of case (i) shows that it holds iff $t\le-2$ and $t \ge 3$. But these are clearly incompatible, so case (i) cannot hold.

Or else, as you observed, it is implicit in the problem that $t \ge 0$, so $t \le -2$ is physically irrelevant.

For case (ii), look first at $-16t-32\le 0$. Rewrite this as $-16t \le 32$, and then $t\ge -2$. Rewrite the condition $t-3\le 0$ as $t \le 3$. So as far as the formula is concerned, everything is OK if $-2\le t\le 3$. But $t \ge 0$, so we conclude that the answer is $0 \le t \le 3$.

Thus we got a complete and correct analysis out of the "doesn't work."

However, let's start again from $-16(t+2)(t-3) \ge 0$.

This is equivalent to $(t+2)(t-3) \le 0$.

That is true if (i) $t +2 \le 0$ and $t-3\ge 0$ OR (ii) $t+2\ge 0$ and $t-3 \le 0$. (A product is $\le 0$ if one term is $\le 0$ and the other is $\ge 0$.)

Case (i) is physically irrelevant. Anyway, it yields the incompatible $t \le -2$ and $t \ge 3$.

Case (ii) yields $t \ge -2$ and $t \le 3$. For physical reasons this should be corrected to $0 \le t \le 3$.

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from your inequality -16*t^2+16*t+128>32 there is one question if at least 32 does it mean that not only >32 but also >=32? so rewrite your equation like this

-16*t^2+16*t+128>=32


-16*t^2+16*t+96>=0 now dive it by -16 and change > by < so we will have t^2-t-6<=0 solution is t1=3 t2=-2 ,but because feet can't be negative answer will be [0 3]

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