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I fail.

Name: Anonymous 2007-09-06 9:58 ID:z+bAOkXc

Fuck. This problem is getting on my nerves. I can't for the life of me figure out something as simple as this... Maybe because ever time I see my physics book I think of choking the teacher but that's a whole other story. I know it's a really dumb, cliche  problem but I have no idea where to start on it so can you guys help me out?
A ball is thrown upwards with an initial velocity of 25m/s and at the same instant another ball is dropped from 15m high. how long will it be before they are at the same height?

Any help?
Yes, I suck cocks at physics.

Name: Anonymous 2007-09-06 10:24 ID:5F2USz+w

ok let's go

What do we know?

a=-9.81

ball thrown upwards
u=25
t=t for 2nd ball
s=0

ball dropped
u=0
t=t for 1st ball
s=15

so fuck V...

We'll be using...
s = ut + att/2

1st ball
0 = 25t + att/2

2nd ball
15 = att/2

All together now...

0 = 25t - 15
15 = 25t
0.6 = t

Name: Anonymous 2007-09-06 10:36 ID:z+bAOkXc

Thanks, dude.
I couldn't find those equations in the book. I guess they expected us to derive them ourselves.
Again, thanks.

Name: Anonymous 2007-09-06 11:09 ID:Heaven

>>3
lol what kind of physics book are you using

Name: Anonymous 2007-09-06 12:59 ID:hrPvxWrb

>>3
You fail pretty hard. Read up on speed-time graphs and the equations...

v = u + at
s= ut + (1/2)at^2
V^2 = U^2 + 2as

Name: Anonymous 2007-09-06 18:22 ID:Heaven

>>3
It's pretty easy to do without the equations as well, but you need (very) basic calculus.

y''[t]=acceleration=-g => y'[t]=v=-gt+y'[0] => y[t]=-gt^2/2+y'[0]t+y[0] where y is a coordinate direction parallel with the gravitational force.

For the dropped ball, y_db[0]=15, y'_db[0]=0
For the thrown ball, y_tb[0]=0, y_tb'[0]=25
And we want to find some t_0 such that y_db[t_0]=y_tb[t_0]

-g*t_0^2/2+15 = -g*t_0^2/2+25*t-0 => t_0=15/25=0.6 s

Name: Anonymous 2007-09-06 18:34 ID:c3KkS3IX

Logic works too.  The ball thrown up will slow down the same exact amount that the ball dropped will speed up.  So, it makes sense that the time it takes for the balls to reach each other is the same exact amount of time it would take the balls to reach each other in the absence of gravity; in otherwords, 25*t=15 so t=0.6s.

Name: Anonymous 2007-09-07 6:18 ID:nCQgm5cA

Heh... None of the rest of my class got it which is weird for a school full of Indian and Asian kids with an average 3.8GPA...
So I asked the teacher to show us how to do it and he spent like 20 minutes trying to get it done but didn't get anywhere close to finishing. A fire drill saved his ass...
It seems like my tuition is going to waste.

Thanks for all the input though.

Name: Anonymous 2007-09-07 12:07 ID:Ot0iF30O

>>8
How old are those in your class? 15-16?

Name: 4tran 2007-09-07 18:56 ID:FBK0H149

>>8
Tuition?  Are you in a private school?  You can probably save yourself tons of money and get a better education in a public school.

Name: Anonymous 2007-09-07 18:58 ID:Heaven

>>10
but private looks better on your college application.

because you pay extra for it.

Name: Anonymous 2007-09-08 4:09 ID:xcUk++Wp

Can't go to a public school because I'm an expat living in a country with a very shitty education system.

Name: Anonymous 2007-09-08 23:29 ID:4rrszP3r

Learn to significant digits...
.60 s

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