Banked Curve Without Friction: Calculating Height at Increased Velocity

[eestep]

Homework Statement

A car is traveling around a circular, banked track without friction. It is initially traveling around bottom of ramp where inner radius of track is 397 meters. Car then increases its velocity by a factor of 1.9. If inclination angle is 13 degrees, what is height (vertical distance above ground) car is traveling at this new velocity? Answer is 239.22 meters.

Homework Equations

v=$$\sqrt{}Rgtan\theta$$

The Attempt at a Solution

v=$$\sqrt{}(397)(9.81)tan(13)$$=29.99

 

[Stonebridge]

You have correctly calculated the velocity of the car initially at the bottom of the ramp.
Now increase that velocity by a factor of 1.9 and calculate the new value of r, using your formula. This will tell you, by trigonometry, how far up the slope the car needs to be.
[are you certain the increase is "a factor of 1.9", it seems a lot; almost double]

 

[eestep]

After calculating R, do I determine height by multiplying it times tan(13)? I am certain but am getting 330.96.

 

[Stonebridge]

Yes. The vertical height up the slope is x tan 13
where x is the increase in the radius.

 

[eestep]

Thank you!