Here's the question you clicked on:
soapia
a fire hose held near the ground shoots water at a speed of 6.5 m/s. at what angle(s) should the nozle point in order that the water land 2.0 m away, why are there two different angles? in the diagram, the hose is held straight up at theta degrees going at 6.5 m/s and with water falling down.. the distance between initial and final water displacement is 2m.. a fire hose held near the ground shoots water at a speed of 6.5 m/s. at what angle(s) should the nozle point in order that the water land 2.0 m away, why are there two different angles?
There are two trajectories a projectile can travel to reach a destination, the high and low orbit. The furthest destination can be obtained with an angle of Theta = 45°. The high and low orbit are obtained with equal angles away from this Theta: Theta1 = 45°+Phi (high orbit) and Theta2 = 45°-Phi (low orbit). formula for the inclined throw \[x _{m}=\left( v _{0}^{2}\sin \left(2.\Theta\right) \over g \right)\] solving for Theta gives Theta = 13,83 ° or Theta = 90°-13,83° = 76,17° (See of it as mirrorred around the 45° axis.
Sorry, I will tell you how I came to that: Water that comes out of a hose are actually only water drops with much the same speed and direction. The beam that the water forms, are only water drops that have a coherence, we call laminar streaming . The water drops that leave the water hose have an initial speed at the nozzle. I make an approximation by supposing the angle between the horizontal at the nozzle and the horizontal 2 meter further being zero, which I suppose I can do for answering this problem (it's like saying that the earth is flat, sort of speak). If the hose is pointed in any upwards direction from the earth's surface, the speed of the water drops decreases until the water drops reach a climax height and return to the earth surface with increasing speed, until the drops join the earth's surface (if you think about this, this speed at dropping point is equal to the initial speed, if it were not for the friction with the air. The speed is slower when the drops of water hit the ground, because they heated the air and themselves by friction) . Now, the paths that all these drops follow can be described by geometric figures, with mathematical equations: they follow the path of a parabola, with a quadratic equation. To analyse and quantise the trajectories I can use the scientific method of Rene Descartes and use an orthogonal axis structure with origin at the point of the nozzle, and the horizontal axis parallel to the ground, and the vertical axis straight upwards to the beautiful sky. We may call the axes X and Y, just like you probably saw somewhere before, but we can use any symbol for naming the axes. (orthogonal means 'straight angle', thus axes have straight angle to one another).
Like this we simplify the three dimensional space into two dimensions X and Y, a plane instead of space. In that plane fit lines that describe equations in two variables x and y. I can already tell that quadratic equations have two (or one) solutions and here this is the case: there are two parabolas that fit into the description of the water trajectories, as solution to your question. Now I will use something you must learn in physics: vectors: adding vectors and splitting vectors in their orthogonal components.(one can also multiply vectors but this is here not needed). With the orthogonal axes structure you can divide speeds (as vectors) into components (and I always used letters X and Y, and Z for space). I have to make another definition: the angle Theta is the angle between the water direction and the horizon, or thus axis X. Now I divide the speed of 6,5 m/s at angle Theta into composing speeds vertical and horizontal (Y and X components)I call: Vy and Vx. What do I know about Vx? It is constant (approximately, so without the friction) and Vx=cos Theta . 6,5m/s (Vx is cosine Theta times 6,5 meter per second). (see of this technique as projecting the speed vector onto the X-axis. and likewise I make Vy = sin Theta . 6,5m/s
I search the time ( t ) needed for the drops to travel from nozzle to ground 2 meter further. This speed is contant (with friction neglected) and I use formula for uniform rectilinear motion: \[V _{x} = \left( s \over t \right) \] (speed v is distance over time, or thus the distance s travelled per unit of time) and thus \[t = \left( s \over V _{x} \right)\] t = 2meter/(6,5m/s cos Theta) this is t in function of Theta Now we observe the second componant Vy: The speed Vy is at t = 0 (when a water drop starts its travel in free fall): Vy = sin Theta . 6,5m/s Sir Isaac Newton was describing motion of bodies by introducing forces acting on them, forces caused by gravition. Now Newton was smart, but his theorie was not correct entirely (see Albert Einstein and his gravitation theory). But the problem that you got, with the hose, can be solved with Newton: Water drops that move upwards fall into the sky, and fall back. They fall back because they are attracted by the other big drop of (partially liquid) mass of rock, called Earth. This earth exerts force on the waterdrops, and they are accelarated with (for this problem almost) constant gravitational acceleration, I call 'g'. The formula that describes this motion is (if you don't know this yet, let me know, I can try to explain it): \[s = s _{0} + v _{0} . t - \left( g.t ^{2} \over2\right)\] with s the distance travelled, s0 the initial distance (is here not relevant, and zero) and v0 is initial speed ( sin Theta . 6,5 m/s), and g the gravitational acceleration, and t the time At t = 0 the distance travelled by a water drop that leaves the nozzle, is zero.
\[v _{0} . t = \left( g.t ^{2} \over 2 \right)\]
The last formula I get by equalling the last two terms of the former equation: then solving for t gives
\[t = \left( 2.v _{0} \over g \right)\]
Why this formula? The last two terms are equal when the waterdrops fall on the ground, and s, distance travelled, is zero (in y direction, since ground level y = 0). If I get time t out of this in function of Theta, I can solve for Theta.
\[t = \left( 2m \over 6,5m/s.\cos \theta \right)\]
\[t = \left( 2.v _{0}\over g \right) = \left( 2.6,5m/s.\sin \theta \over 9,81m/s² \right)\]
equalling t in both equations gives an equation in Theta:
\[\sin \theta . \cos \theta = \left( 2m.9,80665m/s²\over2.6,5m/s.6,5m/s \right)\]
We can write in goniometry with the dubbling formulas: \[\sin \theta . \cos \theta = \left( \sin 2 \theta \over 2 \right)\] and thus have the last formula solved for sin 2 Theta: \[\left( \sin 2\theta \over 2 \right) = \left( 2m.9,80665m/s² \over 2.6,5m/s.6,5m/s \right)\] and get \[\sin 2\theta = 0,464378698\] and to get Theta I take the inverted sin: \[2\theta = \sin^{-1} 0.464378698 = 27,67 °\] Theta is \[\theta =\left( 27,67° \over 2 \right) = 13,835°\]
This is one of the possible angles to get water 2 meter further. The value can differ a bit, because I used an approximation for the gravitational acceleration g. Some use 9,81m/s² and others use 9.80665m/s² . It is not a constant, and varies over different locations on earth. Since I don't know where you live, I took just approximation.
I will do the rest tomorrow.
continued: the last formula \[\sin2θ=0,464378698\] has actually two solutions for theta: you can go and try: the value for theta can be \[90^{0}-13,835^{0} = 76,165^{0}\] This is so because of the symmetries of goniometric functions: \[\sin \left( 2.76,165^{0} \right) = 0,46437\]
So there we have the answer for the question: "Why are two angles possible?"