Finding the time when a rock hits the ocean, if its height above the water is modeled by a quadratic

Question

Question:

A rock is thrown upward from the top of a $112$-foot high cliff overlooking the ocean at a speed of $96$ feet per second. The rock’s height above ocean can be modeled by the equation $H(t) = −16t^2 + 96t + 112.$

When does the rock hit the ocean?

I have attempted in solving it through making the equation $$0 = −16t^2 + 96t + 112$$ as I expected $0$ to be the lowest point (a.k.a. the ocean), but that did not lead to the desired solution.

my attempt Sorry for the bad handwriting

Answer 1

So, everything is fine in your attempt up until this line:

$$t^2 - 6t = 7$$

From here what you did was factor the left-hand side and get

$$t(t+6) = 7 \implies -t^2 = 13 \implies t = \pm \sqrt{13}$$

• The first equation is slightly wrong, in that $t(t+6) = t^2 + 6t$, not $-6t$.
• The implication is wrong as well. I see no clear reason to conclude that $-t^2 = 13$. I guess you added $6$ to the right-hand side. Interpreting the first equation instead as $$t(t-6) = 7$$ I can sort of see where this mistake comes from. However, adding $6$ on the left-hand side does not negate the $6$. You just get $t(t-6)+6 = 7$. If you want to cancel out something with an addition, it has to an entirely separate term all on its own. For instance, if I had $$ 2x^2 - 6 = 7$$ then I can add $6$ to both sides: $$ 2x^2 \underbrace{- 6 + 6}_{=0} = 7+6 \implies 2x^2 = 13 $$
• Can't say I at all see why you would conclude the use of a $-t^2$ instead of a $t^2$ (positive), though.
• Even then, $-t^2 = 13$ means $t^2 = -13$, which results in two not-real numbers for $t$. (If you know about complex numbers, they will be $t = \pm i \sqrt{13}$. Of course, since we're talking about time, this would be a nonsensical answer, and a key hint that something is wrong.)

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Now, as for how to actually take this on.

Keep your terms on the left-hand side:

$$t^2 - 6t - 7 =0$$

This equation is valid and correct. There are two common methods you can use to take this on:

• Method $\#1$: Factoring: For this method, in the general quadratic $x^2 + bx + c$, we want factors of $c$ which add up to $b$. Suppose those numbers are $p,q$. (Thus $p+q=b$ and $pq=c$.) Note that sometimes one or both of $p,q$ must be negative. Then $$ x^2 + bx + c = (x+p)(x+q) $$ and you can solve the equation from there. In this case, you'll want to see that $-7$ and $+1$ are your factors. (Their product is $-7$ and sum to $-6$.) Then $$ t^2 - 6t - 7 = (t+(-7))(t+1)=(t-7)(t+1) =0$$ Setting each factor equal to zero individual yields $t=7$ and $t=-1$. (If you want some review on factoring quadratics, look here on Khan Academy, as a start.)

• Method $\#2$: Quadratic Formula: If you recall, the quadratic formula states $$ ax^2 + bx + c = 0 \text{ if } x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a} $$ Replacing $x$ with $t$, and letting $a=1, b = -6, c = -7$, we can apply this formula. The rest is just straightforward arithmetic, and leads to the same solutions as before. (Some review on the quadratic formula through Khan Academy can be found here.)

(There are certainly many other methods, but these are the main two people tend to use and some of the first people tend to learn.)

Either way, now that we know $t=7$ and $t=-1$, we have one further issue.

Negative time does not make sense in the scenario. (How could the rock hit the water one second before you threw it in the first place?)

Hence, $t=7$ seconds is the final answer.