Calculating time of falling object

1. Sep 10, 2007

anglum

1. The problem statement, all variables and given/known data

A person standing at the edge of a cliff kicks a stone over the edge with a speed of 20m/s. The cliff is 51m above the water. Gravity acceleration is 9.8m/s

how long does it take the stone to fall and hit the water, answer in seconds

2. Relevant equations

D= Vi t + 1/2A tsquared

3. The attempt at a solution

51m = D

Vi = 20m/s

A = 9.8m/s

t= ???

when i solve for that

i get 2.551 = t + tsquared

how do i solve for t at that point?

2. Sep 10, 2007

learningphysics

does he kick the stone horizontally?

3. Sep 10, 2007

anglum

it doesnt say it just says kicks the stone over the cliff at speed of 20m/s

4. Sep 10, 2007

anglum

if he kicks it straight down does my equation i used work?

if he kicks it horizontally what equation is then needed?

5. Sep 10, 2007

learningphysics

I think we are to assume it is kicked horizontally... horizontal velocity has no effect on vertical displacement.

So if it is kicked horizontally... vi = 0 (vertical direction).

Are there other parts to the question?

If it is kicked straight down, then your equation works, but you made an algebra mistake.

6. Sep 10, 2007

anglum

what algebra mmistake did i make... and if it is kicked horizontally why would they give the 20m/s info

7. Sep 10, 2007

anglum

since it is 51m down and gravity is -9.8 does it matter if i do them as negatives or positives in this case?

8. Sep 10, 2007

learningphysics

the coefficient of tsquared here:

seems wrong to me.

I don't know... I'm wondering if there are more parts to the question that use the 20m/s. Or maybe it is a trick question.

9. Sep 10, 2007

learningphysics

Be consistent... if you take up as positive and down as negative, then stick with that... that would mean displacement is -51m. acceleration is -9.8.

You can also choose down as positive, and up as negative if you want... but whatever you choose, be consistent...

10. Sep 10, 2007

anglum

ok solving with 20m/s as the initial velocity

it looks like this

51=20t + 1/2 9.8 tsquared
51=20t + 4.9 tsquared
51/4.9 = 20t/4.9 +4.9tsquared/4.9

10.08= 4.08t + tsquared
2.551 = t + tsquared (then i am stuck and not sure how to solve for this)

or if i solve with initial velcoity at zero i get 3.226 seconds

11. Sep 10, 2007

anglum

does the horizontal velocity have anything to do with the time it would take to fall to the water?

12. Sep 10, 2007

learningphysics

The last line is wrong... It's a quadratic equation, so you can use the quadratic formula.

But I really doubt the 20m/s is straight down...

Last edited: Sep 10, 2007
13. Sep 10, 2007

learningphysics

No it doesn't. Are there other parts to the question?

14. Sep 10, 2007

learningphysics

do they give an angle by any chance?

15. Sep 10, 2007

learningphysics

yes, this is right if the 20m/s horizontal.

16. Sep 10, 2007

anglum

they do not give an angle and the only other part of the question is at what speed does the rock hit the water?

17. Sep 10, 2007

learningphysics

Oh... ok, well this part requires you to know the 20m/s. how would you approach this part?

18. Sep 10, 2007

anglum

ok it was obviously 20m/s horizontal since my answer of 3.226 was correct

now to solve for the speed at impact i would use the equation

Vi = 0 or 20 m/s
A = 9.8 (negative of positive)
D = 51m
solve for Vf???

Last edited: Sep 10, 2007
19. Sep 10, 2007

learningphysics

yup, that's the equation, but don't forget the squares it's: Vf^2 = Vi^2 +2AD

that will give you the vertical velocity at impact... you need to use the horizontal velocity also, and get the magnitude of velocity.

20. Sep 10, 2007

anglum

i need the magnitude of velocity to solve for Vf?

do i use 0 as the intitial velocity or 20?