Question

A ball is thrown straight up in the air at a velocity of 24 feet/second. The height of the ball at t Seconds is represented by the equation h=24t-16t^2. find the maximum height and the time at which it will be there?

can someone walk me through this?

Answer 1

Flvs?

Answer 2

any ways the way you solve this is by they equation f(t) = -16t^2 + vt + s

Answer 3

This should be on the question

Answer 4

It says use the equation that I gave to solve?

Answer 5

I dont see the equation

Answer 6

Please post it so i can help

Answer 7

look at the pic

Answer 8

Ok got it thank you

Answer 9

Go ahead and graph that in something like GeoGebra or Wolfram Alpha

Answer 10

k

Answer 11

You will get the anwers in bot of the because it tells you where the max and min is with a picture of the graph which is really helpfull

Answer 12

Find h'

Answer 13

then set h' = 0. Then solve for t

Answer 14

**what are the white people when you look at viewing ? People not logged in

Answer 15

ok sorry my internet is being slow

Answer 16

If this is algebra (not calculus), you can find the peak of the parabola. It is the vertex of the parabola

Answer 17

so with this i just do the middle term. and yeah its algebra but I couln't figure out the labeling for the vertex equation

Answer 18

write the equation in standard form (x^2 term is first)

Answer 19

\[ h= -16t^2+ 24t + 0 \]
match to
y = a x^2 + b x + c

Answer 20

ok got that

Answer 21

vertex is at t= -b/(2a)

Answer 22

-.75?

Answer 23

a= -16 , b= 24

-b/(2a) is -24/-32 = + 24/32 = ¾ = 0.75
so you were close.

Answer 24

ok

Answer 25

you should notice from the graph that the vertex does not occur at a negative t. that is a clue to double check what you did.

Answer 26

so i thats my line of symmetry?

Answer 27

yes. notice that 0.75 is exactly ½ between the "zeros" at x=0 and x=1.5 See graph

Answer 28

great thanks!