Okay, so, from now on let's assume that g = 10 unless stated otherwise.

Here's a problem:

A faucet is turned on; its spout faces straight downwards. It has radius 0.5 cm, and a 0.5 cm radius stream of water flows out of it at a velocity of 2 m/s. What's the radius of the stream 50 cm below the faucet?

Equating PE + KE per kilogram at the faucet and half a meter below, we have

g * (0.5 m) + 1/2 * (2 m/s)^2 = 1/2 * v^2,

where v is the speed of the water at that point below. Then

v = sqrt( 2 * g * (0.5 m) + (2 m/s)^2)

Then we have the fact that the volume throughput at both the points are (approximately) the same. Then, with r being our radius we're looking for,

pi * (0.005 m)^2 * (2 m/s) = pi * r^2 * v.

This means

r = sqrt((0.005 m)^2 * (2 m/s) / sqrt(2 * g * (0.5 m) + (2 m/s)^2))

With g = 10 N, this gives 0.0037 m. I.e. 0.37 cm.

This answer is not the true answer, even in the 'ideal' case, but it's close enough. In the 'ideal' case, where nothing slows the water down and it's perfectly streamlined, you could still account for the fact that the velocity of the water is not pointing straight downwards.

Here's my question::
A helicopter takes off along the vertical with an acceleration a=3m/s/s and zero initial velocity. After a certain time t1, a bullet is fired from the helicopter. At the point of take off on ground, the sound of the shot is heard at a time t2=30sec after the take off of helicopter. Find the velocity of the helicopter at the moment when the bullet is fired assuming that velocity of sound is c = 320m/s.

2 dimensions? This is a one dimensional problem. What's 'x'? I can't understand what you're doing. But I guess you do...

The way I did it was to solve

t1 + (3 m/s/s) / 2 * t1^2 / (320 m/s) = 30 s

for t1.

oh.. where'd you get that equation? I used 2 dimensions.. the vertical, and horizontal.. where x is the horizontal displacement and y is the vertical displacement.. which are the exact same.

I didn't catch that "At the point of take off on ground, the sound of the shot is heard " untill after I had started using 2d.. works out the same though

That equation comes from adding the time from liftoff to firing the bullet (t1), to the time it takes the sound to reach the ground (the distance the helicopter's traveled divided by 320 m/s).

ah ic.. Should we let someone else post a question.. since all of us have gone already. How about the next person to post (who hasn't already) gets to go ahead and ask a question?

Okay, I'll go. A .2g bullet is ejected from a gun at 500m/s. The bullet then hits a fly. At the time the bullet hits the fly, the fly is 20m away from you. The path of the bullet it perpendicular to the path from you to the fly. The fly weighs .04g and the bullet has constant velocity (assume) before it hits the fly. The collision is inelastic. Three seconds after the fly is hit it screams. Assuming the speed of sound in air is 343m/s how long after the fly screams can you hear it? (Heh, I have to solve this one myself first, it should work out fine).

Okay, I'll go. A .2g bullet is ejected from a gun at 500m/s. The bullet then hits a fly. At the time the bullet hits the fly, the fly is 20m away from you. The path of the bullet it perpendicular to the path from you to the fly. The fly weighs .04g and the bullet has constant velocity (assume) before it hits the fly. The collision is inelastic. Three seconds after the fly is hit it screams. Assuming the speed of sound in air is 343m/s how long after the fly screams can you hear it? (Heh, I have to solve this one myself first, it should work out fine).

So I used the conseration of momentum to find the combined velocity then multiplied by three to fine the displacement after three seconds, then used the pathagorean theorem to find the displacement between me and the fly/bullet, then used the equation v=d/t and solved for t.

It would be directly in front of me, but the bullet would be coming from either the left or the right as it hits the fly. Like watching cars on the highway while you're standing off to the side. Maybe I didn't explain it well the first time.