#1

An object is thrown up at 20.6 m/s and falls over the edge of building. It strikes the ground 5.00s after being thrown. What is the height of the building?

#2

wait,

seriously?

seriously?

#3

This is common sense. Don't help him, guys.

#4

Common sense?

#5

How is it common sense, is it not a parabola?

#6

about 100 feet, I did it in my head so you might want to check it.

EDIT; Sorry more like 94 95 ish feet

EDIT; Sorry more like 94 95 ish feet

#7

dont you just multiply them together?

#8

This is common sense. Don't help him, guys.

n00b!

I should know. But I can't be bothered to work it out... Sorry

#9

no its not common sense. dont be a douche. just write out your formulas, figure out what you know and what you dont know, solve for variables, substitute equations...all that jazz. good luck bro

#10

103 meters.

I failed physics though, someone feel free to correct me.

I failed physics though, someone feel free to correct me.

#11

u have to count acceleration of the ball. initial velocity is 20.6m/s, and acceleration is -9.81m/s^2.

you don't have the final velocity, but the distance traveled is x. total time (displacement) is 5.00s, so u can figure that out.

you don't have the final velocity, but the distance traveled is x. total time (displacement) is 5.00s, so u can figure that out.

#12

not common sense, it requires extensive math skillz and scientific reasoning

i threw away my cheat sheet from last year, and my school was too cheap to give us AP physics. otherwise i'd help you.

i threw away my cheat sheet from last year, and my school was too cheap to give us AP physics. otherwise i'd help you.

#13

Looks n00bs... It's like this:

o

I ^

I I ???

I I

I I

I I at 20.6m/s

I

I

I

I

I

V

???

o

I ^

I I ???

I I

I I

I I at 20.6m/s

I

I

I

I

I

V

???

#14

Its been 2 years since i did applied maths....

Give me a paper and pencil and i'll solve this in 10mins....

But............

Just can't be bothered!!

Do it yourself dude!!!

Yup, thats the way its down....

Upside is +ive and down is -ive...

Use Newton's 3 equations of motion to find the missing unknown...

Give me a paper and pencil and i'll solve this in 10mins....

But............

Just can't be bothered!!

Do it yourself dude!!!

Looks n00bs... It's like this:

o

I ^

I I ???

I I

I I

I I at 20.6m/s

I

I

I

I

I

V

???

Yup, thats the way its down....

Upside is +ive and down is -ive...

Use Newton's 3 equations of motion to find the missing unknown...

#15

You write out and graph an equation. Right?

Seems kinda easy to me. Unless, I'm wrong, and if I am, I'm sorry.

Seems kinda easy to me. Unless, I'm wrong, and if I am, I'm sorry.

#16

Looks n00bs... It's like this:

o

I ^

I I ???

I I

I I

I I at 20.6m/s

I

I

I

I

I

V

???

Interesting thesis...

#17

Yup, thats the way its down....

Upside is +ive and down is -ive...

Use Newton's 3 equations of motion to find the missing unknown...

newtons 3rd wont help at all you dip****. newtons nothing will help. you need the motion equations x=V(i)t+(1/2)at^2 and that stuff

what tards hang out on ug...

#18

don't forget if its thrown up ur gonna have to figure ou how high it goes befor it peaks then - that distance from the total to get ur displacement

#19

divide it into 2 parts. ist part, until the ball hits the hghest point. its final velocity is zero at the highest point

intitial is 20.6. acceleration is -9.8(gravity)u cna now find the time and distance. then you do basically the same for the second part. u=at v sqd = u sqd +2as s=ut + 1/2 at sqd etc etc

intitial is 20.6. acceleration is -9.8(gravity)u cna now find the time and distance. then you do basically the same for the second part. u=at v sqd = u sqd +2as s=ut + 1/2 at sqd etc etc

#20

all you need is the time it takes and how high it goes up first, then use the leftover time and the total height (h+whatever you find) to calculate the rest

#21

18.5m

u= - 20.6

a= 9.8

s = ?

t = 5s

N2L:

S = ut + 0.5at^2

s = -20.5(5) + 0.5(9.8)(5^2)

s = + 18.5m

I think thats only half of the solution.. but whatever.... someone elz can solve the rest!!

u= - 20.6

a= 9.8

s = ?

t = 5s

N2L:

S = ut + 0.5at^2

s = -20.5(5) + 0.5(9.8)(5^2)

s = + 18.5m

I think thats only half of the solution.. but whatever.... someone elz can solve the rest!!

#22

^ Time =/= 5s that's the tricky bit...

The whole process is 5 second...

The whole process is 5 second...

#23

^ Time =/= 5s that's the tricky bit...

The whole process is 5 second...

Note i used u = - 20.6 (- means the body is going upwards)

and g = + 9.8

The time taken for the body to go upwards and come down is 5secs....

Time taken for the - velocity to convert into + velocity has been taken into account...

I've don't applied maths dude... I've solved like a million problems like this... I know how its done... And i just cant be bothered to solve more...

#24

Note i used u = - 20.6 (- means the body is going upwards)

and g = + 9.8

The time taken for the body to go upwards and come down is 5secs....

Time taken for the - velocity to convert into + velocity has been taken into account...

I've don't applied maths dude... I've solved like a million problems like this... I know how its done... And i just cant be bothered to solve more...

When in hell is an upward motion negative? and btw...the answer is 48.12 m. and dont **** with me, im in ap physics

#25

Vo=20.6 m/s, a=-9.8m/s^2, t=5, y=?

y=20.6m/s(5)+1/2(-9.8m/s^2)(5s)^2

y=-19.5m

meaning height is 19.5m?

y=20.6m/s(5)+1/2(-9.8m/s^2)(5s)^2

y=-19.5m

meaning height is 19.5m?

#26

ok theres a discrepancy with this issue. Are u on the ground, or on the top of the building

#27

ok theres a discrepancy with this issue. Are u on the ground, or on the top of the building

Yeh, I'd like to see the actual question, the wording could make a big difference, it seems like there is info missing the way it's worded at the moment.

#28

It's like this:

o

^ - I ^

I I I

I I u=-20.6 (- cuz its in - direction)

I I

I g=+9.8 I I

V I

I

I

I

I I

V + V

By solving the equation as i did above, you get the +ive s (displacement).

i.e. the displacement in the +ve direction.... Which is actually the displacement of the ball from its highest position to the bottom of the ground.

Now find the distance the ball has travelled to reach its highest position.

u = -20.6

v = 0

a = +9.8

s = ?

N3L:

u^2 = v^2 + 2as

(-20.6)^2 = 0 + 2 (9.8)(s)

s = 21.65m

The ball has travelled to reach the top.

Something doesn't feel right!!!!

aah, whatever i can't be bothered!!

I'm in med school now... I don't need maths and physics anymore!!

EDIT: wtf?! why doesn't the diagram come out right in the post???!

o

^ - I ^

I I I

I I u=-20.6 (- cuz its in - direction)

I I

I g=+9.8 I I

V I

I

I

I

I I

V + V

By solving the equation as i did above, you get the +ive s (displacement).

i.e. the displacement in the +ve direction.... Which is actually the displacement of the ball from its highest position to the bottom of the ground.

Now find the distance the ball has travelled to reach its highest position.

u = -20.6

v = 0

a = +9.8

s = ?

N3L:

u^2 = v^2 + 2as

(-20.6)^2 = 0 + 2 (9.8)(s)

s = 21.65m

The ball has travelled to reach the top.

Something doesn't feel right!!!!

aah, whatever i can't be bothered!!

I'm in med school now... I don't need maths and physics anymore!!

EDIT: wtf?! why doesn't the diagram come out right in the post???!

*Last edited by af_the_fragile at Sep 30, 2007,*

#29

Let me elaborate. If your on the top of the building, most of the time will be spent getting back to ground level. Also, angle of tragectory is needed to calculate the x and y components. after that, you can find how tall the building is from the top to the bottom.

If it were from the ground to the building, again angle would be needed to find the trajectory of the building. There would also have to be a distance from the building so that we know when the ball goes above the building.

If it were from the ground to the building, again angle would be needed to find the trajectory of the building. There would also have to be a distance from the building so that we know when the ball goes above the building.

#30

Ok, so you're on top of a building and throw the ball up at 20 m/s and it hits the ground after 5 secs? This is pretty easy, but what are you using for gravity? My class is stupid so we just round to 10 haha.

#31

Let me elaborate. If your on the top of the building, most of the time will be spent getting back to ground level. Also, angle of tragectory is needed to calculate the x and y components. after that, you can find how tall the building is from the top to the bottom.

If it were from the ground to the building, again angle would be needed to find the trajectory of the building. There would also have to be a distance from the building so that we know when the ball goes above the building.

This is not a projectile!

The dude throws the ball up from the edge of the building. The ball goes straight up until its velocity is 0. The it comes straight down and hits the ground below the building.

Use simple Newtons laws of motion to calculate it...

#32

Yeh, I'd like to see the actual question, the wording could make a big difference, it seems like there is info missing the way it's worded at the moment.

I wrote it exactly as written on the worksheet. You are on the top of a building throwing it upwards, and after arcing, it goes straight down to the ground.

Thanks to those that are helping, don't be such idiots to those that are being jerks about it.

#33

Let me elaborate. If your on the top of the building, most of the time will be spent getting back to ground level. Also, angle of tragectory is needed to calculate the x and y components. after that, you can find how tall the building is from the top to the bottom.

If it were from the ground to the building, again angle would be needed to find the trajectory of the building. There would also have to be a distance from the building so that we know when the ball goes above the building.

Lol, we use standard Earth gravity: -9.81 m/s^2

#34

When in hell is an upward motion negative?

When you decide to state that down is the positive direction.

#35

Ok well I got 13.602 meters give or take some rounding. Any idea if this is right? I've only been taking physics for 3 weeks haha but this problem is just like what we've been doing.

#36

This is how you've solve it using Newton's Laws.....

Applied Maths way...

BTW, what grade question is this???

cuz this looks too simple for a university question... If its a university question, i might be doing something wrong...

EDIT:little mistake... answer is 19.8m

Correction:

S = ut + 0.5at^2

s = -20.5(5) + 0.5(9.8)(5^2)

s =

s = 19.5m

EDIT: Assuming you're in high school.... This method is right. All the other graphic methods concerning projectiles or something more complicated would be a little too complex for a high school physics question...

Applied Maths way...

BTW, what grade question is this???

cuz this looks too simple for a university question... If its a university question, i might be doing something wrong...

EDIT:little mistake... answer is 19.8m

Correction:

S = ut + 0.5at^2

s = -20.5(5) + 0.5(9.8)(5^2)

s =

**-103**+ 122.5s = 19.5m

EDIT: Assuming you're in high school.... This method is right. All the other graphic methods concerning projectiles or something more complicated would be a little too complex for a high school physics question...

*Last edited by af_the_fragile at Sep 30, 2007,*

#37

This is how you've solve it using Newton's Laws.....

Applied Maths way...

BTW, what grade question is this???

cuz this looks too simple for a university question... If its a university question, i might be doing something wrong...

Looks right to me.

#38

Woohoo, a chance to practice my rather rusty physics knowledge. I've just started studying it at Uni and can't remember a bloody thing.

Here goes:

initial velocity

acceleration

time

displacement

s = ut + (1/2)at^2

= 20.6 x 5.00 + (1/2) x -9.81 x 5.00^2

= -19.6m

The building is 19.6m tall.

Here goes:

initial velocity

*u*= 20.6ms^(-1)acceleration

*a*= -9.81ms^(-2)time

*t*= 5.00sdisplacement

*s*= ?s = ut + (1/2)at^2

= 20.6 x 5.00 + (1/2) x -9.81 x 5.00^2

= -19.6m

The building is 19.6m tall.

#39

Woohoo, a chance to practice my rather rusty physics knowledge. I've just started studying it at Uni and can't remember a bloody thing.

Here goes:

initial velocityu= 20.6ms^(-1)

accelerationa= -9.81ms^(-2)

timet= 5.00s

displacements= ?

s = ut + (1/2)at^2

= 20.6 x 5.00 + (1/2) x -9.81 x 5.00^2

= -19.6m

The building is 19.6m tall.

Exactly the way i did it....

#40

W00t, thanks a ton guys.

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