I have this problem that I am stuck on, i'm not the best at tackling physics problems.

A firefighter directs a stream of water from a fire hose into the window of a burning building. The window is 20.0m above the level of the nozzle and 60.0m away. The hose produces a jet of water at a constant nozzle speed of v0=30.0m/s. the firefighter can control the angle (theta initial) that the water is aimed at by moving the nozzle with her hands.

(a) the firefighter discovers that there are exactly two angles (initial angles) that she can hold the hose for which the water gets into the window. What are the two angles?

( the fire fighter further discovers that if she backs slowly away from the building there is a distance Dmax where there is only one angle (theta max) which gets the water into the window. If she does any farther back, she cannot get the water into the window at all. Find Dmax and THETAmax.

Instead of considering a constant stream of water, think about throwing a ball into the window instead. You also need to split the problem into vertical and horizontal components. You know the initial speed u, the distance s and the acceleration a in both directions (or you can work them out using a little bit of trig). All you need to do is use the Newtonian equations of motion to work out the horizontal and vertical components of the final speed. I imagine what will happen is that you'll find a quadratic equation for theta that will give you your two solutions for part a. For part b you'll find that the same quadratic equation gives you one (repeated) root for a particular horizontal distance, if the distance is greater than that, the solutions to the quadratic will be complex (the number under the square root in the quadratic formula will be negative).

This is about decomposition of vectors into orthogonal directions. Helpfully, one direction is parallel to the ground and the other is perpendicular, such that gravity acts only in one of the directions.

See if you can finish part one:

Assuming (I have no diagram) that 0 degrees is a direction parallel to the ground and in the direction of the fire, and that 90 degrees is up, then the hose is at (sx0,sy0) = (0, 0) and the initial velocity of the water is
(vx0,vy0) = 30 m/s (cos theta, sin theta) = ( 30 cos theta m/s, 30 sin theta m/s )
The equation of motion in the x direction is
sx = sx0 + vx0 t = 0 + (30 cos theta m/s) t
And since the window is at (Fx, Fy) = ( +60 m, + 20 m), the only time the water could hit the window is when sx = Fx or t = Fx/vx0 = 2/(cos theta) s.

Now the equation of motion in the up direction is:
sy = sy0 + vy0 t - 1/2 g t^2
So for the water to hit the window, we need sy = Fy.

But we need to have a solution in both directions at the same time, and we know what t is.
So we have 20 m = 0 + (30 sin theta m/s) ( 2 /(cos theta) s) - (g/2)( 2 /(cos theta) s)^2

20 m - (60 m)(tan theta) + g (1s)^2 (sec theta)^2 = 0
But (sec theta)^2 = 1 + (tan theta)^2 so using T = tan theta we have:
quadratic equation in T with solutions:

tan theta = (-B + sqrt(B^2 - 4AC))/2A
tan theta = (-B - sqrt(B^2 - 4AC))/2A

Logically this corresponds to an angle where the water hits the window on the way up and a different angle where it hits it on the way down. So if there is a distance where only a single angle hits -- then that must be at the top of the water's vertical movement.

So I get results of between 25 and 30 degrees AND between 75 and 80 degrees, assuming g = 9.81 m/s^2.

Now how would you go about part two?

// Edit -- less spelled out