#1

I have a physics question from review that I forgotten how to do.

A certain physics teacher is frustrated with Mac Computers, so she throws one straight up from the top of a 50 - m tall building. The computer's initial velocity is 20m/s [straight up]. Find (a) the maximum height of the computer, (b) the amount of time it takes to reach that height, (c) the amount of time it takes the computer to reach the ground, and (d) the final velocity of the computer as it hits the ground.

I know it's shameful going to the pit for school questions. But they haven't given out text books in my school yet, so I'm confused.

A certain physics teacher is frustrated with Mac Computers, so she throws one straight up from the top of a 50 - m tall building. The computer's initial velocity is 20m/s [straight up]. Find (a) the maximum height of the computer, (b) the amount of time it takes to reach that height, (c) the amount of time it takes the computer to reach the ground, and (d) the final velocity of the computer as it hits the ground.

I know it's shameful going to the pit for school questions. But they haven't given out text books in my school yet, so I'm confused.

#2

go to a better school.

#3

uvast

#4

a. 15m, b.5 seconds, c. 16 seconds, d. 89

#5

huh, Im doint this exact same thing in physics right now,

gravity is 9.81 and thats the acceleration and I forget what else to do (my physics class is a BS class)

all we do is play with these little dune buggy things, talk, draw pictures and throw crap off the roof

gravity is 9.81 and thats the acceleration and I forget what else to do (my physics class is a BS class)

all we do is play with these little dune buggy things, talk, draw pictures and throw crap off the roof

#6

Maximum height is when the velocity reaches 0 as it's going up.

Each second, it's being pushed down at 9.8 m/s/s.

The rest should be easy enough.

And as I'm sure people have said by the time I finished this post, Math & Science Help Thread.

Each second, it's being pushed down at 9.8 m/s/s.

The rest should be easy enough.

And as I'm sure people have said by the time I finished this post, Math & Science Help Thread.

#7

Hmm,

Okay thanks for the input, I'll post there next time.

Okay thanks for the input, I'll post there next time.

#8

I think I've learned that last year, but I can't remember it anymore. I didn't like my teacher, she couldn't explain things well.

She had big breasts though

She had big breasts though

#9

a. 15m, b.5 seconds, c. 16 seconds, d. 89

Sigh,

don't post random answers ..

#10

Sigh,

don't post random answers ..

#11

dont you have a physics book?

and i know how to do this ****, im just to lazy to look it up.

and i know how to do this ****, im just to lazy to look it up.

#12

I think I've learned that last year, but I can't remember it anymore. I didn't like my teacher, she couldn't explain things well.

She had big breasts though

please, do go on

#13

i still recall this pretty well.

#14

dont you have a physics book?

and i know how to do this ****, im just to lazy to look it up.

I already said in my first post that they didn't hand out textbooks yet

#15

there's some formula u can use....but i forgot it

#16

a = -g = -9.81 m/s^2 not 9.81 m/s^2...

something wierd i found out but acceleration to the ground is negative...

(after being taught my middle school years that thr is no negative accel...god schools so conflicting)

something wierd i found out but acceleration to the ground is negative...

(after being taught my middle school years that thr is no negative accel...god schools so conflicting)

#17

Also, you'll need a couple equations:

V2 = V1 + AT, V2 = Final Velocity, V1 = Starting Velocity, A = Acceleration (Force of Gravity [-9.8 m/s/s on Earth], T = Time.

At max height, V2 will be 0.

0 = 20 + (-9.8)T will get you the time when it's at the max.

I forgot the distance formula. :P I think something like D = 1/2(V1 + V2T) or something. I forgot.

EDIT:

(after being taught my middle school years that thr is no negative accel...god schools so conflicting)

It's not negative. Acceleration is a vector, it has a magnitude (the 9.81), and a direction. The reason it's 0 is that it implies a direction towards the spot it originated. Think of it as 9.81 m/s/s, 90 degrees.

V2 = V1 + AT, V2 = Final Velocity, V1 = Starting Velocity, A = Acceleration (Force of Gravity [-9.8 m/s/s on Earth], T = Time.

At max height, V2 will be 0.

0 = 20 + (-9.8)T will get you the time when it's at the max.

I forgot the distance formula. :P I think something like D = 1/2(V1 + V2T) or something. I forgot.

EDIT:

a = -g = -9.81 not 9.81...

something wierd i found out but acceleration to the ground is negative...

(after being taught my middle school years that thr is no negative accel...god schools so conflicting)

It's not negative. Acceleration is a vector, it has a magnitude (the 9.81), and a direction. The reason it's 0 is that it implies a direction towards the spot it originated. Think of it as 9.81 m/s/s, 90 degrees.

*Last edited by Kapps at Sep 13, 2009,*

#18

Also, you'll need a couple equations:

V2 = V1 + AT, V2 = Final Velocity, V1 = Starting Velocity, A = Acceleration (Force of Gravity [-9.8 m/s/s on Earth], T = Time.

At max height, V2 will be 0.

0 = 20 + (-9.8)T will get you the time when it's at the max.

I forgot the distance formula. :P I think something like D = 1/2(V1 + V2T) or something. I forgot.

ITS (DELTA X) = .50( V i + V f ) x (Delta T)

Delta is the wierd triangle greek symbol tht stands for "change in"

#19

Also, you'll need a couple equations:

V2 = V1 + AT, V2 = Final Velocity, V1 = Starting Velocity, A = Acceleration (Force of Gravity [-9.8 m/s/s on Earth], T = Time.

At max height, V2 will be 0.

0 = 20 + (-9.8)T will get you the time when it's at the max.

I forgot the distance formula. :P I think something like D = 1/2(V1 + V2T) or something. I forgot.

EDIT:

(after being taught my middle school years that thr is no negative accel...god schools so conflicting)

It's not negative. Acceleration is a vector, it has a magnitude (the 9.81), and a direction. The reason it's 0 is that it implies a direction towards the spot it originated. Think of it as 9.81 m/s/s, 90 degrees.

really cause i use like an '05 textbook tht states wht i said...but wht do i know im just a student studying a book.

#20

Alright,

I've managed to figure out (a), and (b)

I can't seem to get the answer for (c) and (d).

As I recall the distance formula was d = 1/2(V1 + V2)T

I've managed to figure out (a), and (b)

I can't seem to get the answer for (c) and (d).

As I recall the distance formula was d = 1/2(V1 + V2)T

#21

ITS (DELTA X) = .50( V i + V f ) x (Delta T)

Delta is the wierd triangle greek symbol tht stands for "change in"

That's the same as what I just said.

T is the change in time, however many seconds passed. The distance you travel is the change in your location, or change in X. Vi is Velocity Initial, Vf is Velocity Final, it's just V2/V3/whatever is easier if you have more than two.

#22

Kinematic equations are your new best friends. Use them.

#23

well i hate physics because the teacher sux and i have to teach myself...all he does is teach by showing us videos of professors at havard teaching physics, pausing the video at times to rephrase wht they guy just said on the video...

#24

Alright,

I've managed to figure out (a), and (b)

I can't seem to get the answer for (c) and (d).

As I recall the distance formula was d = 1/2(V1 + V2)T

Think about it a bit.

You know how long it takes you to reach the top.

Now you have to find out how long it takes from when it reaches the top until it reaches the bottom, then add the time it takes to reach the top.

You know at the top the initial velocity (V1) is 0. You know the acceleration is -9.8 m/s. You know the height is 90.82 (50 m building, thrown (20 / 9.8 * 20) up) metres.

There was another distance formula, but I forgot it. It's used here. ;p

EDIT: D = V1T + 1/2A(T^2). You know D, V1, and A. This is used to find Time. Once you find Time, you know the acceleration. You know the initial velocity. You know the time. You can easily find the final velocity using those.

*Last edited by Kapps at Sep 13, 2009,*

#25

Thanks,

I understand almost everything except that part where you found the total height.

90.82(50 m building, thrown (20/9.8 x 20) )

Why did you do 20/9.8 x 20 ? Don't you just add the building height and the distance from the top of the building thrown to the maximum height ?

EDIT:

Also for that formula you stated, how can I find time when there's two variables for that one. Don't I use the other equation ?

EDIT2:

I figured out how you got 90.82.

But when I input it, I get the wrong answer. ( it states the answer in the question ).

I'm solving for time by using the formula d=1/2(V1+V2)T

I understand almost everything except that part where you found the total height.

90.82(50 m building, thrown (20/9.8 x 20) )

Why did you do 20/9.8 x 20 ? Don't you just add the building height and the distance from the top of the building thrown to the maximum height ?

EDIT:

Also for that formula you stated, how can I find time when there's two variables for that one. Don't I use the other equation ?

EDIT2:

I figured out how you got 90.82.

But when I input it, I get the wrong answer. ( it states the answer in the question ).

I'm solving for time by using the formula d=1/2(V1+V2)T

*Last edited by DGen92 at Sep 13, 2009,*

#26

Thanks,

I understand almost everything except that part where you found the total height.

90.82(50 m building, thrown (20/9.8 x 20) )

Why did you do 20/9.8 x 20 ? Don't you just add the building height and the distance from the top of the building thrown to the maximum height ?

I'm not sure actually. I did something odd, and that part is wrong. Use the maximum height reached. I don't really know what I was trying to do at that part. :p

EDIT:

EDIT:

Also for that formula you stated, how can I find time when there's two variables for that one. Don't I use the other equation ?

There isn't two variables. You know the acceleration (force of gravity). You know the distance (maximum height). You know the initial speed (0). You're trying to find Time. You just plug Time in to two different spots.

EDIT2:

I'm solving for time by using the formula d=1/2(V1+V2)T

You can't unless you already know the final velocity right before it reaches the ground. When it reaches the ground, the velocity isn't 0, it doesn't stop until a moment after it reaches the ground, so assume the velocity is right before it hits the ground.

*Last edited by Kapps at Sep 13, 2009,*

#27

EDIT:

There isn't two variables. You know the acceleration (force of gravity). You know the distance (maximum height). You know the initial speed (0). You're trying to find Time. You just plug Time in to two different spots.

Okay, would you mind explaining how I would rearrange it ?

#28

v(s) = -9.8x+20

I think that is the initial velocity...integrate for the position.

I think that is the initial velocity...integrate for the position.

#29

Okay, would you mind explaining how I would rearrange it ?

D = V1T + 1/2A(T^2)

D = whatever you got for your answer at the maximum height.

V1 = 0. T for the first part doesn't matter, since it's multiplied by 0.

1/2A = -4.9, T^2 = whatever you got for your T.

D / -4.9 = T^2.

Root(D / -4.9) (Distance should be negative in this case, as it's a vector, and you're going down, so you assume it's negative, otherwise you end up with imaginaries) = T. Use the answer that makes sense (aka, the answer that's not a negative amount of time :p).

#30

a) v^2 = v[initial]^2 + 2adI have a physics question from review that I forgotten how to do.

A certain physics teacher is frustrated with Mac Computers, so she throws one straight up from the top of a 50 - m tall building. The computer's initial velocity is 20m/s [straight up]. Find (a) the maximum height of the computer, (b) the amount of time it takes to reach that height, (c) the amount of time it takes the computer to reach the ground, and (d) the final velocity of the computer as it hits the ground.

I know it's shameful going to the pit for school questions. But they haven't given out text books in my school yet, so I'm confused.

0 = (20m/s)^2 + 2(-9.8m/s)(d)

d = (-400m/s)/(-18.6m/s) = 21.5m

b) y = 1/2at^2

21.5m = 1/2(9.8m/s)(t^2)

t = ((21.5m)/(4.9m/s))^1/2

t = 2.1s

OR

v = v[initial] + at

0 = 20m/s + (-9.8m/s)t

t = (-20m/s)/(-9.8m/s)

t = 2.0 s (meh, a little tiny bit off but oh well)

c) Assuming it's not including the time it took to reach the max height.

y = 1/2at^2

50m + 21.5m = 1/2(9.8m/s)(t^2)

t = ((71.5m)/(4.9m/s))^1/2

t = 3.8s

And including the time it took to reach the max height,

3.8s + 2.1s = 4.9s

d) v = v[initial] + at

v = 0m/s + (9.8m/s)(3.8s(remember, the acceleration doesn't start until it reaches the top and starts going back down))

v = 37.2 m/s

#31

D = V1T + 1/2A(T^2)

D = whatever you got for your answer at the maximum height.

V1 = 0. T for the first part doesn't matter, since it's multiplied by 0.

1/2A = -4.9, T^2 = whatever you got for your T.

D / -4.9 = T^2.

Root(D / -4.9) (Distance should be negative in this case, as it's a vector, and you're going down, so you assume it's negative, otherwise you end up with imaginaries) = T. Use the answer that makes sense (aka, the answer that's not a negative amount of time :p).

OH thanks, I fully understand

Great help xD

EDIT:

a) v^2 = v[initial]^2 + 2ad

0 = (20m/s)^2 + 2(-9.8m/s)(d)

d = (-400m/s)/(-18.6m/s) = 21.5m

b) y = 1/2at^2

21.5m = 1/2(9.8m/s)(t^2)

t = ((21.5m)/(4.9m/s))^1/2

t = 2.1s

OR

v = v[initial] + at

0 = 20m/s + (-9.8m/s)t

t = (-20m/s)/(-9.8m/s)

t = 2.0 s (meh, a little tiny bit off but oh well)

c) Assuming it's not including the time it took to reach the max height.

y = 1/2at^2

50m + 21.5m = 1/2(9.8m/s)(t^2)

t = ((71.5m)/(4.9m/s))^1/2

t = 3.8s

And including the time it took to reach the max height,

3.8s + 2.1s = 4.9s

d) v = v[initial] + at

v = 0m/s + (9.8m/s)(3.8s(remember, the acceleration doesn't start until it reaches the top and starts going back down))

v = 37.2 m/s

Thanks, though I already understand it. But that's good to check my answers again.

*Last edited by DGen92 at Sep 13, 2009,*

#32

That's not entirely true. That would be essentially assuming uniform velocity while it's decelerating, which is obviously untrue. It can't be going 20m/s for the amount of time it takes it to go from 20m/s to 0m/s.Think about it a bit.

You know how long it takes you to reach the top.

Now you have to find out how long it takes from when it reaches the top until it reaches the bottom, then add the time it takes to reach the top.

You know at the top the initial velocity (V1) is 0. You know the acceleration is -9.8 m/s. You know the height is 90.82 (50 m building, thrown (20 / 9.8 * 20) up) metres.

The formula given earlier: x = 1/2(v + v[initial])t works here.

t = (v - v[initial])/a

t = about 2.0s

x = 1/2(v + v[initial])t

x = 1/2(20m/s)(2.0s)

x = around 20m.

*Last edited by grampastumpy at Sep 13, 2009,*

#33

That's not entirely true. That would be essentially assuming uniform velocity while it's decelerating, which is obviously untrue. It can't be going 20m/s for the amount of time it takes it to go from 20m/s to 0m/s.

The formula given earlier: x = 1/2(v + v[initial])t works here.

t = (v - v[initial])/a

t = about 2.0s

x = 1/2(v + v[initial])t

x = 1/2(20m/s)(2.0s)

x = around 20m.

Yeah, I realized that after he replied to it, where I said I was wrong. :p

#34

Oh, my bad, lol. If you already found it then woops, if not, then well, there it is.