How Is the Stone's Trajectory Calculated When Thrown at an Inclined Roof?

[postfan]

Homework Statement

We are standing at a distance d=15 m away from a house. The house wall is h=6 m high and the roof has an inclination angle β=30 ∘. We throw a stone with initial speed v0=20 m/s at an angle α= 44 ∘. The gravitational acceleration is g=10 m/s2. (See figure)

(a) At what horizontal distance from the house wall is the stone going to hit the roof - s in the figure-? (in meters)

(b) What time does it take the stone to reach the roof? (in seconds)

Homework Equations

The Attempt at a Solution

I set up 2 equations :
s+15=20*cos(44)*t
s/sqrt(3)+6=20*sin(44)*t-5t^2
and got s=.99 and t=1.11. What did I do wrong?

 

[Simon Bridge]

Homework Statement

We are standing at a distance d=15 m away from a house. The house wall is h=6 m high and the roof has an inclination angle β=30 ∘. We throw a stone with initial speed v0=20 m/s at an angle α= 44 ∘. The gravitational acceleration is g=10 m/s2. (See figure)

(a) At what horizontal distance from the house wall is the stone going to hit the roof - s in the figure-? (in meters)

(b) What time does it take the stone to reach the roof? (in seconds)

Homework Equations

The Attempt at a Solution

I set up 2 equations :
s+15=20*cos(44)*t
s/sqrt(3)+6=20*sin(44)*t-5t^2
and got s=.99 and t=1.11. What did I do wrong?

How did you arrive at those equations?
(Please show your reasoning.)
How did you account for the slope of the roof?

 

[postfan]

I used the equation x=v_0*t+.5*a*t^2. I created a 30-60-90 triangle , and realized that for each distance s horizontally it goes up by s/sqrt(3). The total distance horizontally is 15+s and vertically is (s/sqrt(3)_6) and we are given the acceleration and launch velocity/angle.

 

[Simon Bridge]

As a description, that leaves a lot to be desired.

You need to know where the trajectory parabola of the stone intersects the line of the roof.
What is your strategy for figuring that out?