Can anyone answer this with the steps of Assumption > Given > Formula > Solution?

The height in feet of a bottle rocket is given by the function $h(t)=160-16t^2$, where $t$ is the time in seconds. How long will it take for the rocket to return to the ground? What is the height of the rocket after 2 seconds?

• I see no formula in your post. Further, what did you do to find answer? As it stands, it seems you're just here to ask some homework question without putting in any effort. If you have some function $h(t)$ that gives the height as function of $t$, then what should the value of $h(t)$ be when it is at the ground? – Kenneth Goodenough Aug 1 '15 at 23:48
• You seem to have forgotten to include the function that gives the height of the rocket; please edit your question to include this and some context (motivation, your attempt, etc.) as described here: meta.math.stackexchange.com/questions/9959/… – coldnumber Aug 1 '15 at 23:56
• If the rocket is at the ground, we have $h = 0$, so we have to solve $h(t) = 160 - 16t^2 = 0$. Rearranging gives $160 = 16t^2$, so that $t = \sqrt{10} \approx 3.16$. Inputting in $t = 2$, we find $h(2) = 160-16\cdot 2^2 = 160-64 = 96$. Where exactly in these steps did you get stuck? – Kenneth Goodenough Aug 2 '15 at 0:12
• So if I were to use this on a quadratic formula. What would be the value of c? Would it be zero? Should I transpose the zero? – Lanz Joshua Aug 2 '15 at 0:17
• At time 0 the bottle is at the maximum height. It descent at the rate of 16 feet for the first second, accelerating as it goes down. After $t^2$ seconds equal 10 it reaches the ground (h(t)=0). The $\sqrt{10}$ is the number of seconds at which the bottle will hit the ground. – Moti Aug 2 '15 at 3:25

Maybe you started by asking for something like this...

ASSUMPTION. We start at time $t = 0$ and consider values $t \ge 0.$ We end when the rocket is on the ground, that means height $0,$ so we must have $h(0) = 0$. We need to find $t$ such that $h(t) = 0.$

GIVEN. $h(t) = 160 - 16t^2$. The rocket must start at its highest point because $h(0) = 160$ and the height $h(t)$ only decreases from there on for larger values of $t$.

FORMULA. From some Comments, I think you want to use the quadratic formula to solve the equation to find $t$ such that $h(t) = 0.$ The equation starts as $0 = 160 - 16t^2.$ To use the quadratic formula it needs to be in the form $at^2 + bt + c = 0.$

So that's $-16t^2 + 0t + 160 = 0,$ where $a = -16,\,b=0,\,$and $c = 160.$

SOLUTION: Then the quadratic formula becomes

$$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{0 \pm \sqrt{0 - 4(-16)(1600)}}{2(-16)} = \frac{\pm \sqrt{10240}}{-32} = \mp 3.162278.$$

Of the two values $t = -3.1623$ and $t = 3.1623,$ only the positive one is useful, because we've assumed from the start that $t \ge 0.$ So the rocket crashes to the ground after 3.1623 seconds. This is the same as $\sqrt{10},$ as mentioned in the Comments.

Note: I can understand how someone who considers himself 'not very good at math' would want the security of using the quadratic formula--which always works--even if it turns out to have put you through a little messy arithmetic in this case. I hope you take a good look at what @Kenneth was saying in his Comments. That is a simpler way to solve this particular quadratic equation (in which $b = 0$). Technically, the answer there is really $\pm\sqrt{10},$ but the $\pm$ wasn't mentioned in the Comments because everyone was assuming that $t \ge 0$ from the start.

If I remember my history of math correctly it must have been a bit over 300 years ago that Isaac Newton was first playing around with this and related equations. I'll bet he would have been willing to pay a huge price for your hand calculator.

As discussed in the Comments, the other part is to give the height at $t = 2$ sec. That's just $t(2) = 160 - 16(2^2) = 160 - 64 = 96$ feet high.

I don't know how good you are at graphing and interpreting graphs, but I've put one below of the curve $h(t) = 160 - 16t^2$ for $t$ between $0$ and $\sqrt{10} = 3.1623.$ Orange lines show that the height is $96$ feet at time $2$ seconds. 