# Homework Help: Minimum Velocity to become airbourne

1. Apr 28, 2013

### mouthwash

I dont have a specific question, but was wondering how I would go about solving something like this:
How would you calculate the minimum velocity that a bike, for example, would require to become airbourne off a circular ramp, if you are given say the radius of the ramp. Would you need more info than that?

I thought maybe you could use V = (rg)^0.5
But am hesitant about this because I think thats only to be used for the top of a loop, not the side of a ramp.

2. Apr 28, 2013

### haruspex

Can you be a bit clearer regarding the scenario? Is this going over a hill that's a circular arc?

3. Apr 28, 2013

### mouthwash

basically a biking moving on a perfectly circular ramp or "arc". Like the ones you see in skate parks. How would you calculate the velocity required at the top of the ramp (or if it were a full circle/loop the side) so that the bike would keep moving up ( go air bourne) instead of falling to the ground.

4. Apr 28, 2013

### haruspex

Still not clear. It's part of a circle, not a whole one, and it's an arc in the vertical plane. How much of a full circle is it? Semicircle? If it has any move energy than that needed to reach the top of the ramp it will become airborne - and then fall to the ground. I don't get what distinction you're making.

5. Apr 28, 2013

### mouthwash

In a skate park, there are ramps. The ramps are quarter circles. if a biker goes up the ramp to the top of the ramp, if he isnt going fast enough he will just roll back down, however if he is going fast enough he will keep going up. Basically his velocity out numbers the gravity pulling him back down.
Dunno how else to really explain it.

6. Apr 28, 2013

### haruspex

You can calculate the speed needed to reach the top of the quadrant. (Just energy energy conservation.) Any higher speed will take him off the end of the ramp, but vertically, so in principle he still returns by the same path.

7. Apr 29, 2013

### CWatters

Apply conservation of energy...

The KE at the bottom (= 0.5 m v2) is converted to PE on the way up the ramp (=mgh).

0.5mv2=mgh

mass cancels

v = sqrt(2gh)