Solve for θ and Time: Projecting a Particle at 30 m/s

[|mathematix|]

Homework Statement

A particle is projected at 30 m/s from the foot of a plane which is inclined at 30^o to the horizontal. The particle hits the plane at an angle perpendicular to the plane.
Find the value of the angle of projection θ and the time taken to reach there.

Homework Equations

x=30tcosθ , and y=30tsinθ-5t^2 , where (x,y) are the coordinates of the point of impact and θ is the angle of projection, inclined with the horizontal.

The Attempt at a Solution

Here is a diagram: http://i.imgur.com/FfTwTD7.png
So if the projected particle hits the hill at 90^o then y/x=1/√3 and the velocity will be at 120^o to the horizontal so y'/x'=-√3
I solved these two equations simultaneously and I got θ=-60 but this looks wrong because it doesn't make sense.
Please help :)

 

[tiny-tim]

hi |mathematix|!

So if the projected particle hits the hill at 90^o then x/y=1/√3 and the velocity will be at 120^o to the horizontal so x'/y'=-√3

shouldn't |y'| be larger than |x'| ?

 

[|mathematix|]

Yes, I fixed it :(

 

[tiny-tim]

is your answer ok now?

if not, show us your full working, so we can see what went wrong

 

[|mathematix|]

I did the working using the correct values but I got -60 degree for theta. I can't post the full solution now because it would take time and I have to study english now so I will re do it and see if I get a better answer.
I basically substituted the equations of motion into y/x=1/√3 and y'/x'=-√3 and then solved them simultaneously, found theta and substituted back to find the time, is that the correct approach?

 

[tiny-tim]

I basically substituted the equations of motion into y/x=1/√3 and y'/x'=-√3 and then solved them simultaneously, found theta and substituted back to find the time, is that the correct approach?

yes!