1
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ 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
2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.