Question

A stone was dropped off a cliff and hit the ground with a speed of 120 ft/s. What was the height of the cliff?

Answer 1

use the formula \[V_2^2 = V_1^2 + 2ah\]
where
V_1 is initial speed
V_2 is final speed
a is acceleration
h is height

since it was dropped, initial velocity is 0
final speed is 120
acceleration is 9.8 (because of gravity)

so just substitute
\[\implies (120)^2 = 0^2 + 2(9.8) h\]
solve for h

does that help?

Answer 2

Thats a physics based equation right? this is in my calc book and the teacher called me out once on using physics making me use calc to do it. I know it sounds crazy but any ideas?

Answer 3

i see. interesting

Answer 4

yea trust me I want to do it with projectial motion equation myself but she is not taking it

Answer 5

well..we know that the derivative of distance is speed...and the derivative of speed is acceleration

are you studying differential or integral calculus?

Answer 6

Integral, so does that mean I take the double integral of 120?

Answer 7

if you take the double integral of that you get 0 lol

Answer 8

you need a function...

Answer 9

alright how do we find that?

Answer 10

ugh i just wrote something and it got lost =_=

Answer 11

oh man :(

Answer 12

i tried taking the double integral of that equation above. do you think it'll work?

Answer 13

wouldn't that make it 120 -> 120x -> 60x^2 making it the normal tejectory of a projectile?

Answer 14

and would it let me solve for the height?

Answer 15

no it would let me solve for time and i wouldn't be able to figure out the height with out physics 8(

Answer 16

once you know the time you can solve for distance

Answer 17

hmm you make a good point about that integral..i was thinking derivatives

Answer 18

so what would x be?

Answer 19

x would be time

Answer 20

but you dont have time...

Answer 21

i should have put a t

Answer 22

\[\int\limits_{0}^{120} dv=\int\limits_{0}^{t _{f}} adt\]
\[120-0=a*(t _{f}-0)\]
\[\int\limits dv = \int\limits a*dt= v = a*t + c\]

Answer 23

\[\int\limits dx =\int\limits v*dt\]
\[\int\limits\limits dx =\int\limits\limits (a*t+c)*dt\]
\[x =(a*t^2+c*t+c _{2})\]

Answer 24

awesome man I think i got it from here

Answer 25

you figured out how to solve for the constants?

Answer 26

no i thought i did for a sec but it didn't work out

Answer 27

v = 0 at t=0
x=0 at t=0
so all of our constants are zero

Answer 28

i screwed up earlier
\[x=(\frac{1}{2}*a*t)\]

Answer 29

oh man i'm wicked confused now

Answer 30

ok, let's back up and solve for our constants of integration using v =0 at t=0, and x=0 when t=0

Answer 31

we solved for this earlier
\[v=at+c\]

Answer 32

now we can substitute our initial condition that v=0 when t=0
(0)=a*(0)+c

Answer 33

0=0+c
c=0

Answer 34

so v=at

Answer 35

does that make sense?

Answer 36

yup that sounds good