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

Question

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?

Answer 1

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.