Question

a catapult lunches a boulder with an upword velocity of 112ft/sec. the height of the boulder 'h' in feet after 't' second is given by the function h = -16t^2+ 112t +30. how long does it take the boulder to reach its maximum height?what is the boulder maximum height? round to the nearest hundredth if necessary.

Answer 1

any one can you help to solve this?

Answer 2

Okay, do you recognize what sort of equation h = -16t^2 + 112t + 30 is?

Answer 3

it is qudratic one

Answer 4

okay, what sort of curve does it make?

Answer 5

i think it should be upword

Answer 6

parabola, hyperbola, circle, ellipse?

Answer 7

ok

Answer 8

don't think of the curve as the trajectory of the boulder, this is a plot of height vs. time...

Answer 9

it's a parabola.

Answer 10

thank you

Answer 11

does this parabola open upward (looks like a bowl) or does it open downward (upside down bowl)?

Answer 12

hint: the problem asks for maximum height :-)

Answer 13

maximum height intial velocity is 0

Answer 14

where did you get the idea that the initial velocity is 0?

Answer 15

before launch, it is 0, but at the time of launch, it is 112 ft/s, upward

Answer 16

that's represented by the 112t term in the equation

Answer 17

the -16t^2 term represents the effects of gravity accelerating the boulder toward the ground. the +30 term represents the height at the time of launch

Answer 18

physics

Answer 19

ok

Answer 20

yeah, but you don't actually need to know any physics to solve this problem. just a little algebra and analytic geometry

Answer 21

because the problem asks for the maximum height, this must be a downward opening parabola. if the parabola opens upward, there is no maximum height, right?

Answer 22

i thinks when parabola is downword it is maximum ya?

Answer 23

yes, the maximum will be at the vertex of the parabola. do you know how to find the vertex from that equation?

Answer 24

please show me

Answer 25

okay, there are a number of ways to do it. what class are you taking?

Answer 26

college algebra 1

Answer 27

ple show me the easyest way

Answer 28

okay, then we won't use calculus :-)

so to find the vertex of a parabola we can rearrange the formula into "vertex form"
which is y = a(x-h)^2 + k

with the formula in vertex form, we can read off the coordinates of the vertex directly: it is simply (h, k)

Answer 29

here we have

height = -16t^2 + 112t + 30

we need to rearrange that as

height = -a(t - h)^2 + k

any idea how to do that? have you heard of completing the square?

Answer 30

ya but need help

Answer 31

(b/2)^2

Answer 32

is the formula?

Answer 33

first thing to do is factor out the -16

height = -16(t^2 - (112/16)t - 30/16)

then b = -112/16 and b/2 = -56/16 = -7/2 and (b/2)^2 = 49/4

agreed?

Answer 34

sure

Answer 35

thank you

Answer 36

that gives us
height = -16(t-7/2)^2 +16*(49/4)+30 or
height = -16(t-7/2)^2 + 226

Answer 37

ok

Answer 38

h = 7/2
k = 226

so the vertex is at (7/2, 226) or (3.5, 226)

Answer 39

thank you very much

Answer 40

a different approach is to remember that if you have a parabola in standard form
y = ax^2 + bx + c (which we do, without any rearrangement)
then the x-coordinate of the vertex is at x = -b/(2a)
and the y-coordinate of the vertex is just the function evaluated at that value of x

so we will do it that way as well, as a check on our first solution:

height = -16t^2 +112t + 30

a = -16
b=112
c = 30

vertex x-coordinate = - b/(2a) = -112/(2*-16) = -112/-32 = 56/16 = 7/2 = 3.5
vertex y-coordinate = -16(3.5)^2+112(3.5) + 30 = -16(12.25) + 112(3.5) + 30 = 226

Answer 41

ok

Answer 42

Here's a graph:

Answer 43

thank you

Answer 44

i got it now thank you very much for your great help

Answer 45

Glad I could help! Thanks for the medal, you just promoted me a rank :-)

Answer 46

you help me a lot thanks for your idea have a good night

Answer 47

you too!