Calculating time of falling object

[anglum]

Homework Statement

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

Homework Equations

D= Vi t + 1/2A tsquared

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?

 

[learningphysics]

does he kick the stone horizontally?

 

[anglum]

it doesn't 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]

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]

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]

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]

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...

 

[learningphysics]

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]

do they give an angle by any chance?

 

[learningphysics]

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]

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?

 

[learningphysics]

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]

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]

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

 

[learningphysics]

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

exactly.

 

[anglum]

so my answer would be 37.4057m/s for the velocity upon impact?

 

[learningphysics]

so my answer would be 37.4057m/s for the velocity upon impact?

yeah, looks good to me.

 

[anglum]

thanks a lot ... i am takin this class online so its really a tough concept to grasp that way... i will prob be askin for more help later on... if i get stuck that is...

 

[learningphysics]

thanks a lot ... i am takin this class online so its really a tough concept to grasp that way... i will prob be askin for more help later on... if i get stuck that is...

no prob. yeah, that is a tough way to learn this stuff... but you're doing really well... keep posting whenever you need help.