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Use calculus to determine how long it takes the sphere to reach its maximum height, also determine what the maximum height is.

The original question is A life form standing on the surface of an unknown planet throws a small inanimate sphere vertically upwards and then steps backwards. The life form releases the sphere at a height of 1m; after 4 seconds the sphere has reached a height of 37m, after a further 4 seconds the sphere is at a height of 25m.

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    $\begingroup$ Can you tell us where you are stuck? $\endgroup$ – tpb261 May 29 '14 at 8:21
  • $\begingroup$ I dont know how to figure out how long the sphere will reach its maximum height $\endgroup$ – user152246 May 29 '14 at 8:28
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    $\begingroup$ Obviously...What have you tried so far? $\endgroup$ – stochasm May 29 '14 at 8:34
  • $\begingroup$ Duplicate of math.stackexchange.com/questions/810000/… (which currently has no upvoted answer). $\endgroup$ – Gerry Myerson May 29 '14 at 9:21
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The idea here is that, after the sphere is released, the only force notably acting on it is the planet's gravity, which is effectively constant near the planet's surface. In other words, if we let $h(t)$ indicate the height (in meters) of the sphere above the planet's surface $t$ seconds after the life form has released it, then $h''(t)$ is constant. From there, we can show that $h(t)=at^2+bt+c$ for some constants $a,b,c.$ Using the facts that $h(0)=1,$ $h(4)=37,$ and $h(8)=25$ will give us a system of equations that allow us to solve for $a,b,$ and $c.$

Can you take it from there?

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  • $\begingroup$ I got the answer for that, I'm just confused as to how I would use that information to find how long it will take for it to reach its maximum height $\endgroup$ – user152246 May 29 '14 at 17:27
  • $\begingroup$ Well, if you have a differentiable function on a closed interval, where can a maximum occur? At endpoints and at points where...what? $\endgroup$ – Cameron Buie May 29 '14 at 17:31

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