Question

With what initial velocity must an object be thrown upward (from ground level) to reach the top of the Washington Monument (approximately 550 ft.)? i have to use integrals to solve and assume the a(t)= -32 feet per second Help working please? i already know the answer

Answer 1

assuming a(t) = k

what is the integral wrt-t ?

Answer 2

i have to take the integrals of acceleration and velocity to get the answer

Answer 3

i know, so lets start by integrating a(t) = k, wrt-t

Answer 4

i've already got acceleration integrated to -32t +C

Answer 5

good, but lets revise that some

int a(t) = k -> v(t) = kt + v , we know at t=0, we want to know the initial velocity "v"

now integrate it again v(t) = kt + v , wrt-t to get a function that determines height

Answer 6

i already have height tho... and i don't do well with letters lol... sorry..but i have tried taking the integral of velocity and i got a position equation but i can't figure out how to use it because i have two variables

Answer 7

well, show me what you came up with :) im sure your just looking at it a little bit off is all

Answer 8

yeah i think so..its giving me a headache..

Answer 9

can i talk to you on skype that way i can just send a picture?

Answer 10

i dont do skype

to follow my idea:

int: v(t) = kt + v -> h(t) = kt^2/2 + vt + h , such that when t=0, height=0 giving us h=0

h(t) = kt^2/2 + vt


filling in for when h(t)=550, and k=-32

550 = -32t^2/2 + vt

Answer 11

i can't follow that...i don't understand what your substituting the variables for..

Answer 12

k is some constant of acceleration, they defined it for us as -32

v is some initial velocity (velocity starts with a "v") that is to be determine.

and h is some initial height (height starts with an "h") which we know to be at the ground at a time of zero seconds, so h=0

Answer 13

the question is really, at what time is the velocity equal to zero?

Answer 14

oh okay..i'm representing height with s(t)

Answer 15

yeah..but to find that do i need to find the velocity at 550ft?

Answer 16

-32t + v = 0 when v = 32t, agreed?

Answer 17

idk i have never seen that formula

Answer 18

the velocity at the height of the building is when it has slowed down all the way and is starting to fall again. so the velocity at the top of the building is at zero

Answer 19

yes i have that written..i meant to say time..sorry..lol

Answer 20

its your integration of acceleration

a(t) integrates to v(t) = -32t + C , where C is the intial velocity that I simply like to call "v"


when is the velocity function v(t) equal to zero? well, -32+v = 0 when ... v = 32t: why? (-32t+32t=0)

Answer 21

so, lets determine our height at v = 32t

\[550=-16t^2+vt\]
\[550=-16t^2+(32t)t\]

Answer 22

but i thought you would have to plug in zero to time before you sub. velocity for anything..

Answer 23

.... might be able to do this in one shot if i keep my head on straight

solve v(t) = 0 for t :)

-32t + v = 0 when t = v/32 might be more prudent

Answer 24

\[550=-16t^2+vt\]
\[550=-16\frac{v^2}{(-32)^2}+v(\frac{v}{-32})\]
and solve for v

Answer 25

you dont need to sub solve the velcoity equation in order to eliminate one of the variables in the height equation

Answer 26

either in terms of t = v, or v = t and then do some proper substitutions

Answer 27

either way, say we go with the first set up; v = 32t

550 = -16t^2 + 32t^2
550 = 16t^2
550/16 = t^2
sqrt(550/16) = t

since v(t) = -32t + v

0 = -32(sqrt(550/16)) + v

v = 32(sqrt(550/16))

Answer 28

this is how i have been taught tho..i don't understand exactly how you are getting to -32t + 32t=0

Answer 29

is it true that we want to determine when the velocity is equal to zero?

Answer 30

t wouldnt be squared in that second step tho..

Answer 31

no..im looking for the initial velocity....which will be a big number because of how far the object is traveling

Answer 32

is it true that we want to know when the velocity function equals out to zero?

Answer 33

big is relative

Answer 34

the initial velocity is your +C value of your velocity function

Answer 35

it will be close to 187.617

Answer 36

http://www.wolframalpha.com/input/?i=32%28sqrt%28550%2F16%29%29

i know

Answer 37

yeah..but i can't plug it in because i don't have it..

Answer 38

ugh...this is frustrating

Answer 39

you have 2 equations in 2 unknowns, which can be solved by substitution

Answer 40

\[550=-16t^2+Ct\\0=-32t+C\]

let C = 32t

\[550=-16t^2+(32t)t\]
\[550=-16t^2+32t^2\]
\[550=16t^2\]
\[\frac{550}{16}=t^2\]
\[\sqrt{\frac{550}{16}}=t\]
since C=32t\[C=32\sqrt{\frac{550}{16}}\]

Answer 41

OR, we could have solved the second equation for "t" instead of "C"
\[550=-16t^2+Ct\\0=-32t+C\]

let t = -C/32, and sub it into the first equation
\[550=-16(-\frac{C}{32})^2+C(\frac{C}{32})\]
\[550=\frac{16C^2}{32^2}+\frac{C^2}{32}\]
\[550=\frac{16C^2}{32^2}+\frac{32C^2}{32^2}\]
\[550=\frac{48}{32^2}C^2\]
\[550\frac{32^2}{48}=C^2\]
\[\sqrt{550\frac{32^2}{48}}=C\]

Answer 42

opps, i dropped a negative

Answer 43

\[550=\frac{-16C^2}{32^2}+\frac{32C^2}{32^2}\]
\[550=\frac{16}{32^2}C^2\]
thatll be better

Answer 44

okay..my instructor is telling us to plug in the zero to t in the problem V(t)= -32t + c
and then that zero carries through and the problem just tells me that c is initial velocity...but i already know that..so i understand what your doing but i'm not sure if thats what i need to do

Answer 45

it is what you need to do; define C in terms of t, or define t in terms of C either way ... and sub it in properly to solve

Answer 46

so don't plug in the zero to t..just replace v and represent v with zero?

Answer 47

correct, if i read you right

we know that at t=0 we have the initial velocity that carries up into the height equation.

we want to determine a substitution such that the velocity equation is equal to 0 at some undetermined value of t (or some underdetermined value of C)

Answer 48

our options are therefore:\[0=-32t+C~:~C=32t\]OR\[0=-32t+C~:~t=\frac{C}{32}\]
so the choice is up to you as far as how you want to substitute this into the height function

Answer 49

personally, i think the C = 32t is nicer to play with :)

Answer 50

yes it is..i did that..and i understand i think..let me see if i can get the answer

Answer 51

k, good luck ;)

Answer 52

i got it=) thank you so much.! sorry if i caused any frustration..lol

Answer 53

'sok, i needed the practice ;)

Answer 54

hahaha have a great day;)