A fire hose held near the ground shoots water at a speed of 6.5 m/s

1. At what angle(s) should the nozzle point in order that the water land 2.5 m away?
-I believe the answer is 18,72

2. Sketch the trajectory for the 18; angle
-Please give me exact (x,y coordinates)
-I believe I need 5 pts on the graph

3. Sketch the trajectory for the 72; angle.
-Please give me exact (x,y coordinates)
-I believe I need 5 pts on the graph

To answer the first question, we can use the principles of projectile motion. The water from the fire hose follows a curved path, forming a trajectory. In order for the water to land 2.5 m away, we need to find the angle(s) at which the nozzle should be pointed.

1. We need to consider the horizontal and vertical components of the water's velocity.
Let's call the angle at which the nozzle is pointed as θ and the initial velocity of the water as v = 6.5 m/s.

2. The horizontal component of velocity (v_x) is given by v_x = v * cos(θ).
The vertical component of velocity (v_y) is given by v_y = v * sin(θ).

3. The time of flight (t) is the time it takes for the water to land.
We can find it using the formula: t = 2 * v_y / g,
where g is the acceleration due to gravity (approximately 9.8 m/s²).

4. The horizontal displacement (x) is given by x = v_x * t.

5. Setting x = 2.5 m, we can solve for the angle(s) θ that satisfies the equation.
Rearranging the equation from step 4, we have: θ = arccos(x / (v * t)).

Now let's calculate the angles for the nozzle direction:

θ = arccos(2.5 / (6.5 * t))
Substituting the value of t from step 3, we find:
θ = arccos(2.5 / (6.5 * (2 * v_y / g)))

Now we can calculate the angles.

θ = arccos(2.5 / (6.5 * (2 * v * sin(θ) / g)))
This equation is transcendental, which means it cannot be solved analytically. However, we can use numerical methods or approximation to find the solutions. Based on your belief that the answer is 18° and 72°, let's assume those angles are the solutions.

To check if these angles are correct, we'll substitute their values back into the equation, and see if the distance is indeed 2.5 m.

θ = 18°:
v_x = v * cos(18°)
= 6.5 m/s * cos(18°) ≈ 6.18 m/s

v_y = v * sin(18°)
= 6.5 m/s * sin(18°) ≈ 1.87 m/s

t = 2 * v_y / g
= 2 * (1.87 m/s) / 9.8 m/s² ≈ 0.38 s

x = v_x * t
= 6.18 m/s * 0.38 s ≈ 2.35 m

The distance x is approximately 2.35 m, which is close to 2.5 m but not exactly. Hence, we need to find a more accurate solution.

Now, let's solve the equation numerically using an iterative method like the bisection method or Newton's method to find the exact angles that satisfy the equation.