Question

An object is launched at 19.6 m/s from a height of 58.8 m. The equation for the height (h) in terms of time (t) is given by h(t) = -4.9t2 +19.6t + 58.8. What is the object's maximum height?
A. 58.8
B. 194.04
C. 78.4
D. 117.6

Answer 1

-4.9t2 +19.6t + 58.8. do i divide the equation by -4.9? @zepdrix

Answer 2

forgot to add then i use quadratic formula

Answer 3

no, you differentiate the equation and find out h'(t) = 0, then substitute found t value into h(t)

Answer 4

This isn't calculus level math.

Answer 5

They probably want you to put the parabola in standard form, that way you can identify the vertex.

Answer 6

Ya dividing by -4.9 is probably the right way to start :)

Answer 7

\[\Large h(t) = -4.9(t^2+?t+?)\]

Answer 8

So what do you get for those other two values when you factor the -4.9 out?

Answer 9

ok I get
\[t ^{2}-4t-12\]

Answer 10

a(t^2-4t-12) like that right?

Answer 11

\[\Large h(t) = -4.9(t^2-4t-12)\]Ya looks good so far :)

Answer 12

ok What would be the next step?

Answer 13

We need to turn the t terms into a perfect square.\[\Large h(t) = -4.9(\color{orangered}{t^2-4t}-12)\]

Answer 14

So what value should we add to complete the square?

Answer 15

i use
\[(\frac{ b }{ 2 })^{2}\]

Answer 16

right?

Answer 17

yes

Answer 18

just 2 or is it 2a?

Answer 19

We're going to be a little sneaky.
We're going to do all of this inside of the brackets.
So let's ignore the a for now.

Answer 20

ok

Answer 21

i get 4

Answer 22

so we should add 4 to complete the square?\[\Large h(t) = -4.9(\color{orangered}{t^2-4t+4}-12)\]Ya that sounds right!
We have a small problem though, we can't simply add 4 willy nilly.
What can we do to keep this equation balanced?

Answer 23

ummm....not sure goona take a guess do we add to the other side too?

Answer 24

Good, that's what we would normally do, right?
We would add to BOTH sides to keep things balanced.

There is another trick you should be aware of though:
~We can also keep the equation balanced by adding 4 AND subtracting 4.
If we do that, we're keeping it balanced while at the same time changing the way it looks.

Answer 25

\[\Large h(t) = -4.9(\color{orangered}{t^2-4t+4}-4-12)\]We're adding 4 to complete our square. We'll also subtract 4 to keep the equation balanced.
I colored in orange the terms we'll need for our perfect square.

Answer 26

Darn we shouldn't had taken the -4.9 out of the 58.8 earlier.
It gives us an extra step now. I forgot about that. :)
No big deal though.

Answer 27

Confused on the add/subtract thing?

Answer 28

NO makes sense

Answer 29

Let's combine the -4 and -12.\[\Large h(t) = -4.9(\color{orangered}{t^2-4t+4}-16)\]From here, we really only want the orange part in the brackets.
So let's multiply -16 and -4.9 to get it out of the brackets.

Answer 30

ok soo would it be \[h(t)=78.4(t^{2}-4t+4)\]

Answer 31

Woops, that shouldn't change our a value.
It will be added on the outside.\[\Large h(t) = -4.9(\color{orangered}{t^2-4t+4})+(-4.9)(-16)\]\[\Large h(t) = -4.9(\color{orangered}{t^2-4t+4})+78.4\]

Answer 32

ohh ok

Answer 33

then ill find perf square of the orange part?

Answer 34

You write it as a square* yes.

Answer 35

adrn sep gtg thanks for help ill finish it later :D

Answer 36

It will simplify down to:\[\Large \left(t+\frac{b}{2}\right)^2\]

Answer 37

oh ok :)