# Ball rolls of an edge and lands on and inclined plane

1. Mar 19, 2008

### sinisterguy

1. The problem statement, all variables and given/known data

the ball rolls at a constant speed, fast enough to travel straight off the ledge and eventually lands on the inclined plane. The task is to derive an equation for d as a function of v (the initial horizontal speed of the ball) and theta.

2. The attempt at a solution
I started by trying to find y by using $$\Delta d_{y}= 1/2g \Delta t^2$$
then I moved on to x which is simply $$v \Delta t$$
I also know that $$\Theta = tan^{-1} (\frac{1/2 g \Delta t^2}{v \Delta t})$$
this is all great, but i wasn't quite sure how to get rid of the t

after some more fiddling i also found that $$\Delta t = \frac{d cos \Theta}{v}$$, but along that same train of thought, if $$d = \sqrt{x^2 + y^2}$$ and $$x = d cos \Theta$$ then that wouldn't work.

my teacher told me i wasn't on the right track so i started over, but i haven't gotten anywhere with that. some help to point me in the right direction would be great

2. Mar 19, 2008

### Tom Mattson

Staff Emeritus
I would tackle it using conservation of energy. Such an analysis would be completely time-independent from the get go.

3. Mar 19, 2008

### sinisterguy

we haven't learned about conservation of energy yet (well, not enough to be able to apply any mathematical solution to a problem). so that might be a challenge

4. Mar 19, 2008

### Tom Mattson

Staff Emeritus
OK, then do it using only time-independent kinematic equations. Have you studied projectile motion? If so, then you should have seen an equation for the trajectory of a projectile that doesn't contain t. It is the equation of a parabola that opens downward. You also can write down the equation of the line that contains the inclined plane (remember that the slope of a line is equal to the tangent of its angle of inclination). So basically you can reduce this whole problem to the geometrical problem of finding where the parabola intersects the line.