Solving for Distance d Up Slope of 1/5

[Lancelot59]

I'm given that a particle launched at 30 degrees, and 40m/s from the horizontal axis travels a certain distance d up a constant slope of 1/5.

I need to determine how far up that slope the particle goes.

I got a velocity vector the obvious way:

[tex]\vec{v}=(40cos(30},40sin(30)-9.8t)[/tex]

Then integrating to get position:
[tex]\vec{r}=(40cos(30)t,40sin(30)t-9.8\frac{t^{2}}{2})[/tex]

I'm fairly certain I did this bit correctly. The issue I'm having here is that I don't know how to now solve for the arbitrary distance d that the particle travels up the hill.

I was thinking of writing a function for the slope, then seeing where it intersects the parabolic path.
I got [tex]y=\frac{1}{5}x[/tex] for the slope function. However this is in terms of position x and not time. I'm thinking I can use the x position from the parabolic path as the x parameter in the slope function and then solve the resulting equation. Does this make sense?

 

[tiny-tim]

Hi Lancelot59!

… I'm thinking I can use the x position from the parabolic path as the x parameter in the slope function and then solve the resulting equation. Does this make sense?

Yes.

 

[Lancelot59]

It worked! I was off of the book's answer by 0.1. Thanks.