Question

A projectile is thrown upward so that its distance above the ground after t seconds is given by the function h(t) = -16t2 + 640t. After how many seconds does the projectile take to reach its maximum height?

Answer 1

I can help

Answer 2

thank you

Answer 3

i can help

Answer 4

Have you seen this? \(-\dfrac{b}{2a}\)

Answer 5

It will reach maximum height when
h'(t) = -32t + 640 = 0
-32t = -640
t = -640/-32 = 20 seconds

Answer 6

thank you, could you explain how you did that? @tdunmyer

Answer 7

i asked my cousin the question and he told me to put what he said he's a math teacher

Answer 8

We should probably first determine if the Calculus can be used. If this is Algebra 2 or Trig, we need a different solution.

Answer 9

this is pre calculus @tkhunny

Answer 10

oh okay. well thank you! @tdunmyer

Answer 11

Okay, now answer my first question. Have you seen, \(-\dfrac{b}{2a}\)? If not, we'll have to derive it.

Answer 12

it looks familiar, but I don't really know what it is @tkhunny

Answer 13

It's a neat little idea that can be derived several ways.

h(t) = -16t2 + 640t

a = -16
b = 640

-b/(2a) = -640 / (2*(-16)) = -640/(-32) = 20 <== Same answer as the calculus version.

It tells you the value of 't' for which the height is maximum. It does NOT tell you the maximum height. You must evaluate the function to see that.

You can find -b/2a by
1) Completing the Square
2) Finding the average of the two quadratic solutions.
3) Using a derivative from calculus.
4) There may be other ways.

Here's #2 If we use the quadratic formula, we get for f(t) = 0 \(t = \dfrac{-b\pm\sqrt{Stuff}}{2a}\).

Find the average of those 2

\(\dfrac{\dfrac{-b - \sqrt{Stuff}}{2a} + \dfrac{-b + \sqrt{Stuff}}{2a}}{2} = \dfrac{\dfrac{-2b}{2a}}{2} = -\dfrac{b}{2a}\).

If you can find where the projectile is on the ground, there should be two places, the average of those two places is the where the maximum occurs.