Question

A rock is dropped off the edge of a cliff and its distance (in feet) from the top of the cliff after t seconds is s(t)=16t^2. Assume the distance from the top of the cliff to the water below is 1024ft.

When will the rock strike the water?

Answer 1

@jim_thompson5910

Answer 2

you would solve 16t^2 = 1024 for t

Answer 3

what do you get

Answer 4

so... 16(1024)^2...

Answer 5

no you're solving 16t^2 = 1024 for t

not plugging in t = 1024

Answer 6

O...8

Answer 7

yep, it will take 8 seconds

Answer 8

So how do i find the average velocity when the time interval is [7,8]?

Answer 9

use the following

AROC = (f(x2) - f(x1))/(x2 - x1)

Answer 10

x1 = 7
x2 = 8

Answer 11

can u do a walkthrough?

Answer 12

s(7) = ???

Answer 13

s(t)=16t^2

s(7)=16(7)^2

s(7) = ???

Answer 14

49

Answer 15

16(7)^2 = 16(49) = 784

Answer 16

s(7) = 784

Answer 17

what is s(8)

Answer 18

1024

Answer 19

av=240?

Answer 20

AROC = (s(x2) - s(x1))/(x2 - x1)


AROC = (s(8) - s(7))/(8 - 7)


AROC = (1024 - 784)/(8 - 7)


AROC = (240)/(1)


AROC = 240


so you got it

Answer 21

Thanks so much for your help tonight, hope your not tired of me :)

Answer 22

you're welcome and no I like helping, so you're fine