I have an interesting calc question here but im not sure how to solve it. Can someone perhaps give me a helping hand or guide me through steps?

A balloon that takes images of the earth is shot up in the sky with rockets from 0 ft off the ground is given by the height of the function s(t)= $-18t^2+120t$ .

a) Find velocity after 2 and 4 seconds as it approaches space. b) When does the balloon reach full altitude? c) When does it touch back down to earth?

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    $\begingroup$ You're missing a few things. You write that a balloon is given by a function, but maybe you mean the height of a balloon is given by a function. You have $18^2$ where you might mean $18t^2$. A rock enters the question and then leaves with no effect - what is that about? Please edit your question into something sensible. $\endgroup$ – Gerry Myerson Jun 25 '12 at 12:16
  • $\begingroup$ @GerryMyerson What do you mean what would you add to this question to make sense of it? $\endgroup$ – soniccool Jun 25 '12 at 12:17
  • $\begingroup$ Did you read any of what I wrote, other than the last sentence? I told you about three things that don't make sense, that need your attention. Please, do something about them. $\endgroup$ – Gerry Myerson Jun 25 '12 at 12:19
  • $\begingroup$ Oh i see thank you so much $\endgroup$ – soniccool Jun 25 '12 at 12:20
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    $\begingroup$ $s(t) =-18^2 + 120t$ suggests that the balloon will never come back. On the other hand $-18t^2+120t$ makes more sense. $\endgroup$ – Saurabh Jun 25 '12 at 12:21

Well, I believe that the velocity of an object is its derivative.

So, find the derivative of your initial function.

$$\ s(t) = -18t^2 + 120t $$

With a simple application of the power rule, we arrive at

$$\ s'(t) = -36t + 120 $$

$$or~ v(t) = -36t+120$$

Now, you have a function that will give you the balloon's velocity at any time t, since the derivative is the instantaneous rate of change(aka velocity).

  • $\begingroup$ How would i find when the balloon reaches full altitude and when it touches back down to earth? $\endgroup$ – soniccool Jun 25 '12 at 12:32
  • $\begingroup$ See my comments elsewhere. $\endgroup$ – Gerry Myerson Jun 25 '12 at 12:55

The rate of change of the objects height with time is the first derivative of the function s(t) this also (intuitively) would be the vertical velocity of the balloon.

When the balloon has a zero velocity it's reached the top of its "curve" - its maximum height.

So if $s(t)=-18t^2+120t$ then $s'(t)=-36t+120$ now we must solve this for zero to find its maximum height or evaluate it at 2 or 4 for its velocity after 2 or 4 seconds (I'll leave that last one for you, it's trivial).


when the balloon "touches" back down to earth (more aptly it crashes) the height is zero

$$s(t)=0=-18t^2+120t$$ Right away this tells us that at time 0 the balloon is already at zero height. We're more interested in later on, so we divide by $t$ (knowing it's not zero anymore) then we get. $$s(t)=-18t+120\\18t=120\\t=\frac{120}{18}=\frac{20}{3}$$

In doing physics questions it's useful to write down your knowns (quantities that are given to you) and the things you want to solve for. Looking at this find a way to rewrite the question in a more "step-by-step" way before you solve. Physics questions hone a very useful set of problem solving skills and enrich you with more than just a basic understanding of Newtonian Mechanics - they train you in a type of thinking that is ubiquitous and necessary in many fields of study.

  • $\begingroup$ Also, it's generally good form to mark questions as "homework" if they are. $\endgroup$ – damagedgods Jun 25 '12 at 13:08
  • $\begingroup$ The only thing you've left for OP to do is to remove the apostrophes from "it's curve" and "it's maximum height". $\endgroup$ – Gerry Myerson Jun 25 '12 at 13:12
  • $\begingroup$ @GerryMyerson Fixed, cheers. $\endgroup$ – damagedgods Jun 25 '12 at 13:14

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