# Calculating time of falling object

#### 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?

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#### learningphysics

Homework Helper
does he kick the stone horizontally?

#### anglum

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

#### anglum

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

if he kicks it horizontally what equation is then needed?

#### learningphysics

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

if he kicks it horizontally what equation is then needed?
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.

#### anglum

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

#### 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?

#### learningphysics

Homework Helper
what algebra mmistake did i make... and if it is kicked horizontally why would they give the 20m/s info
the coefficient of tsquared here:

2.551 = t + tsquared
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.

#### learningphysics

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

#### 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

#### anglum

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

#### learningphysics

Homework Helper
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)
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...

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#### learningphysics

Homework Helper
does the horizontal velocity have anything to do with the time it would take to fall to the water?
No it doesn't. Are there other parts to the question?

#### learningphysics

Homework Helper
do they give an angle by any chance?

#### learningphysics

Homework Helper
or if i solve with initial velcoity at zero i get 3.226 seconds
yes, this is right if the 20m/s horizontal.

#### 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?

#### learningphysics

Homework Helper
they do not give an angle and the only other part of the question is at what speed does the rock hit the water?
Oh... ok, well this part requires you to know the 20m/s. how would you approach this part?

#### 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

Vf squared= Visquared +2AD?

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

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#### learningphysics

Homework Helper
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

Vf = Vi +2AD?

Vi = 0
A = 9.8
D = 51m
solve for Vf???
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.

#### anglum

i need the magnitude of velocity to solve for Vf?

do i use 0 as the intitial velocity or 20?

#### learningphysics

Homework Helper
i need the magnitude of velocity to solve for Vf?

do i use 0 as the intitial velocity or 20?
0. That will give you vf... the vertical velocity at impact. What is the horizontal velocity at impact?

#### anglum

im not sure how to calculate horizontalvelocity... and then how do i combine the vertical with the horizontal to get the actual velocity at impact

I calculated the vertical velocity to be 31.61 m/s

#### learningphysics

Homework Helper
im not sure how to calculate horizontal velocity... and then how do i combine the vertical with the horizontal to get the actual velocity at impact
horizontal velocity doesn't change... so it is still 20m/s. because there is no horizontal acceleration.

at the end you need to use the pythagorean theorem to get the velocity at impact.

#### anglum

so im assuming then i take20m/s squared + 31.61m/s squared = the real velocity at impact squared?

#### learningphysics

Homework Helper
so im assuming then i take20m/s squared + 31.61m/s squared = the real velocity at impact squared?
exactly.

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