A ball is thrown into the air with an initial velocity of $16 ft/s$. its height after $t$ seconds is given by $f(x) = 16t-4t^2$ . After how many seconds does the ball reach its maximum height?

I can't remember back whether this fell under the derivative section or the Limit section, therefore i am stuck

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    $\begingroup$ Are you sure the equation is right? From what I see its max will be at t=0 $\endgroup$ – Matt Dec 2 '15 at 2:07
  • $\begingroup$ Sorry, I had a - infront of the 16. my bad. $\endgroup$ – Jon Roy Dec 2 '15 at 2:09
  • $\begingroup$ @Matt - You may be right, however i got two. $\endgroup$ – Jon Roy Dec 2 '15 at 2:16
  • $\begingroup$ Yeah, the formula changed, in the first version it was just 16, not 16t - that or I missaw. 2 is right, get it by setting the derivative to 0 $\endgroup$ – Matt Dec 2 '15 at 2:17

My advice is not to think of problems as being in "sections", but instead to think about what the problem is asking for.

That function is fairly well-behaved - no weird corners or jumps or holes.

So, where it reaches its maximum, it's going to be flat before falling back down, right?

HINT: If it's flat at its maximum what does that tell you about it?

  • $\begingroup$ I calculated the maximum at x = 2, using the increasing/decreasing interval method, seem correct? $\endgroup$ – Jon Roy Dec 2 '15 at 2:12
  • $\begingroup$ @Jon: You got it! $\endgroup$ – Deusovi Dec 2 '15 at 2:20

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