Question

A stone is thrown straight up from the edge of a roof, 675 feet above the ground, at a speed of 18 feet per second.
A. Remembering that the acceleration due to gravity is -32 feet per second squared, how high is the stone 6 seconds later?
B. At what time does the stone hit the ground?
C. What is the velocity of the stone when it hits the ground?

Answer 1

this is easy

Answer 2

i need help setting it up

Answer 3

i understand anti derivatives and finding constants… setting up the problem is the hard part

Answer 4

we can set this up using integration if you want

Answer 5

let us begin with the -32 ft/sec

Answer 6

that would be v(t)?

Answer 7

it's up to you, I would set this up using s(t)
you can use v(t) or a(t) it doesn't matter

Answer 8

ok, just tell me how you would do it ;)

Answer 9

list the following
-32 as your rate of fall
18 as your initial velocity
675 as your initial position

Answer 10

integrate -32 in terms of time for me

Answer 11

don't you think I know what I am doing, annas?

Answer 12

physics nazi!! haha its ok i understand

Answer 13

fine fine s^2 geez

Answer 14

so… s(t) = 16x^2

Answer 15

srry 16t^2

Answer 16

damn... you're skipping steps

Answer 17

v(t) = 32t

Answer 18

a(t) = 32

Answer 19

\[\int\limits -32 dt=32t+C\]

Answer 20

constants! right..

Answer 21

so v(18) = 32(18) + C1 = -32(18)?

Answer 22

so c = 18/16

Answer 23

-18/16

Answer 24

huh? don't you put zero where the t's are?

Answer 25

wait, wouldn't it be (-32t + c) though?

Answer 26

\[s'(t) = \int\limits s''(t) = \int\limits\limits\limits -32dt=-16t+C \rightarrow s'(0)=-16t+C=18\]

Answer 27

\[ s'(0)=-16t+C=18 \rightarrow -16(0) + C = 18 \rightarrow C= 18\]

Answer 28

so a(t) = -32
v(t) = -16t + c1
s(t) = -8t^2 +c1(t) +c2
thats all i understand

Answer 29

okay, we can do that you wanna change from a to v to s

Answer 30

yes plz

Answer 31

so we just solved your c1 as 18

Answer 32

v(t) = -16t + 18

Answer 33

then let us integrate v(t) to find s(t)

Answer 34

so s(t) = -8t^2 + 18t +C2

Answer 35

\[s(t) = \int\limits v(t) = \int\limits (-16t+18) dt = \int\limits -16t dt + \int\limits 18dt \]

Answer 36

yes yes

Answer 37

lead coefficient should still be negative, yes?

Answer 38

\[s(t) = \frac{ -16t^2 }{ 2 }+ 18t + C_2 \rightarrow s(0) = -8t^2+18t+C_2=675\]

Answer 39

so c2 = 675. got it

Answer 40

\[C_2 = 675 \rightarrow s(t) = -8t^2 +18t+675\]

Answer 41

so that is your position function

Answer 42

s(t) = -8t^2 + 18t + 675
A). s(6) = -8(6)^2 + 18(6) + 675

Answer 43

yeah pretty much

Answer 44

math, skull

Answer 45

applications of integration

Answer 46

why isn't this answer correct :(

Answer 47

who said it isn't?

Answer 48

it say how HIGH, so it is asking for the position at t= 6

Answer 49

then at what time does the stone hit the ground
now your position function should be equals to zero then solve for t

Answer 50

ok i got it :) thank you!!

Answer 51

s(6) = -8(6)^2 + 18(6) + 675 = 495 ft

Answer 52

s(t) = -8t^2 + 18t + 675 = 0
use QUADRATIC EQ or any tricks you know to solve for t

Answer 53

wanna do it together?

Answer 54

i got B now

Answer 55

how do you get C?

Answer 56

t = -8.1292 and t = 10.379
pick the positive

Answer 57

well, how do you find the velocity when s(t) = 0?

Answer 58

you use the velocity v(t)
that means you take the derivative of s(t)
but since you already know that you just use it

Answer 59

would i plug v(10.379)?

Answer 60

ya ya

Answer 61

-8(10.379)^2 + 18(10.379) = -674.967128

Answer 62

boom! 100%

Answer 63

thank you!

Answer 64

therefore it is -674.967128 ft/s^2

Answer 65

XOXO

Answer 66

that stone is going to turn into diamond

Answer 67

piece of cake, yeh?

Answer 68

are we still doing the same problem?

Answer 69

you can close it now