Physics help please

[ParadiceCity9]

ugh i don't think my answer's right..

i got 28.4

[Herakilla]

Vf^2 = Vo^2 + (2ax)

Vf^2 = 0 + 303.8

Vf = 17.4

thats how to get Vf right from the data you gave me

EDIT: 9.8 is positive remember

[ParadiceCity9]

Vf^2 = Vo^2 + (2ax)

Vf^2 = 0 + 303.8

Vf = 17.4

thats how to get Vf right from the data you gave me

and what is Vf?

[Herakilla]

Vf is final velocity

[pimpdave]

fucksakes.

I wish I had internets fora when I was in high school. I just had to figure it out for myself, back in the dark ages of the early 90s.

[ParadiceCity9]

what about using vvat?

v = v-naught +at

[ParadiceCity9]

what about using vvat?

v = v-naught +at

that's giving me the answer i got before.

[ParadiceCity9]

I have a feeling, though, that I have to somehow account for the initial velocity, horizontal. the question in the book has the only difference of the initial V as being 11.4 instead of 11...and the answer is 20.8

[Herakilla]

what about using vvat?

v = v-naught +at

ive never heard of that one im taking AP physics and it sounds like your on the first chapter for us wince that was what we did then

[Herakilla]

wow they are just giving you the initial velocity, run my same equation but replace Vo with 11.4 and youll get 20.8. i just tried it right now. apparantly they mean initial velocity in the X direction already

[ParadiceCity9]

wow they are just giving you the initial velocity, run my same equation but replace Vo with 11.4 and youll get 20.8. i just tried it right now. apparantly they mean initial velocity in the X direction already

can you tell me exactly what you plugged in?

[Herakilla]

Vf^2 = 11.4^2 + (2*9.8*15.5)

solve for Vf

[pimpdave]

wow they are just giving you the initial velocity, run my same equation but replace Vo with 11.4 and youll get 20.8. i just tried it right now. apparantly they mean initial velocity in the X direction already

can you tell me exactly what you plugged in?

Just do your own goddamn homework, Cartman, you fatass.

[ParadiceCity9]

Vf^2 = 11.4^2 + (2*9.8*15.5)

solve for Vf

oh oh...that's v^2 = v-naught^2 + 2-delta-x*a

sweet. thanks

[ParadiceCity9]

A swimmer, capable of swimming at a speed of 2 m/s in still water (i.e. the swimmer can swim with a speed of 2 m/s relative to the water), starts to swim directly across a 2.3-km-wide river. However, the current is 0.94 m/s, and it carries the swimmer downstream. How far downstream will the swimmer be upon reaching the other side of the river?


is the answer 1069.5 m?

[Herakilla]

close 1081

X = 1/2(Vo + Vf)(T)

2300 = 2T

T = 1150 seconds

1150*.94 = 1081 m

[ParadiceCity9]

close 1081

X = 1/2(Vo + Vf)(T)

2300 = 2T

T = 1150 seconds

1150*.94 = 1081 m

hm must have typed .93..

[ParadiceCity9]

A horizontal rifle is fired at a bull's-eye. The muzzle speed of the bullet is 686 m/s. The barrel is pointed directly at the center of the bull's-eye, but the bullet strikes the target 0.027 m below the center. What is the horizontal distance between the end of the rifle and the bull's-eye?

i'm pretty sure the time is .0742 seconds. that's what i have so far..

[InkL0sed]

I had this shit down almost a year ago now. I even made a NetLogo program that imitated a bouncing ball in a vacuum.

Hm... I know there's a way to get that online... I should ask blake about that, maybe I can show it off

[ParadiceCity9]

is the answer 50.2229374 m?

[Herakilla]

yes you got the right time, then all you have Vf = 686, T = .0742 and Vo = 0

use those to find X

(x = 1/2(Vf + Vo)(T)) hint hint

[ParadiceCity9]

sweet.

next problem

At some airports, there are speed ramps to help passengers get form one place to another. A speed ramp is a moving conveyor belt that you can either stand or walk on. Suppose a speed ramp has a length of 102 m and is moving at a speed of 2.3 m/s relative to the ground. In addition, suppose you can cover this distance in 76 s when walking on the ground. If you walk at the same rate with respect to the speed ramp that you walk with respect to the ground, how long does it take for you to travel the 102 m using the speed ramp?

haven't started yet.

[ParadiceCity9]

yes you got the right time, then all you have Vf = 686, T = .0742 and Vo = 0

use those to find X

(x = 1/2(Vf + Vo)(T)) hint hint

that gives me exactly 1/2 of what i got. ugh...

I used X = Xo + Vxot to find it.

[ParadiceCity9]

sweet.

next problem

At some airports, there are speed ramps to help passengers get form one place to another. A speed ramp is a moving conveyor belt that you can either stand or walk on. Suppose a speed ramp has a length of 102 m and is moving at a speed of 2.3 m/s relative to the ground. In addition, suppose you can cover this distance in 76 s when walking on the ground. If you walk at the same rate with respect to the speed ramp that you walk with respect to the ground, how long does it take for you to travel the 102 m using the speed ramp?

haven't started yet.

i got 27.4 s

did 102/72 to get m/s for him. then added that to 2.3 since it's the same direction..then divided that into 102.

[Herakilla]

ya thats what i got